Search: keyword:new
|
|
|
|
1, 2, 3, 8, 11, 5, 7, 15, 24, 6, 18, 4, 19, 10, 14, 30, 21, 12, 20, 34, 38, 16, 33, 25, 23, 42, 13, 37, 48, 17, 53, 45, 9, 50, 58, 28, 83, 36, 61, 29, 55, 79, 40, 62, 22, 65, 49, 54, 66, 27, 69, 41, 59, 70, 32, 35, 73, 47, 39, 74, 60, 26, 44, 87, 78, 93, 63, 98, 52, 64, 43, 101, 77, 86, 102
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Conjectured to be a permutation of the natural numbers (and if so, -1 will never appear).
|
|
LINKS
|
Table of n, a(n) for n=1..75.
|
|
CROSSREFS
|
Cf. A361321, A000469, A361323, A336957, A360519, A361102.
|
|
KEYWORD
|
nonn,new
|
|
AUTHOR
|
Scott R. Shannon, Rémy Sigrist, and N. J. A. Sloane, Mar 11 2023
|
|
STATUS
|
approved
|
|
|
|
|
|
|
1, 2, 3, 12, 6, 10, 7, 4, 33, 14, 5, 18, 27, 15, 8, 22, 30, 11, 13, 19, 17, 45, 25, 9, 24, 62, 50, 36, 40, 16, 121, 55, 23, 20, 56, 38, 28, 21, 59, 43, 52, 26, 71, 63, 32, 79, 58, 29, 47, 34, 87, 69, 31, 48, 41, 105, 97, 35, 53, 61, 39, 44, 67, 70, 46, 49, 76, 77, 51, 54, 84, 89, 57, 60, 96
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Conjectured to be a permutation of the natural numbers (if not then there exists a k such that A000469(k) does not appear in A361321).
|
|
LINKS
|
Table of n, a(n) for n=1..75.
|
|
CROSSREFS
|
Cf. A361321, A000469, A361324, A336957, A360519, A361102.
|
|
KEYWORD
|
nonn,new
|
|
AUTHOR
|
Scott R. Shannon, Rémy Sigrist, and N. J. A. Sloane, Mar 11 2023
|
|
STATUS
|
approved
|
|
|
|
|
A360663
|
|
a(n) is the least integer m >= 3 such that n is a centered m-gonal number.
|
|
+0
0
|
|
|
3, 4, 5, 6, 7, 8, 3, 10, 11, 4, 13, 14, 5, 16, 17, 3, 19, 20, 7, 22, 23, 4, 25, 26, 9, 28, 29, 3, 31, 32, 11, 34, 35, 6, 37, 38, 13, 4, 41, 7, 43, 44, 3, 46, 47, 8, 49, 5, 17, 52, 53, 9, 55, 56, 19, 58, 59, 4, 61, 62, 3, 64, 65, 11, 67, 68, 23, 7, 71, 12, 73, 74, 5, 76, 77, 13, 79
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
4,1
|
|
LINKS
|
Table of n, a(n) for n=4..80.
Eric Weisstein's World of Mathematics, Centered Polygonal Number.
Wikipedia, Centered polygonal number.
|
|
EXAMPLE
|
a(16) = 5 since 16 is a centered pentagonal number, but not a centered square or centered triangular number.
|
|
MATHEMATICA
|
seq[len_] := Module[{s = Table[0, {len}], c = 0, k = 3, n, ckn}, While[c < len, n = 2; While[(ckn = k*n*(n - 1)/2 - 2) <= len, If[s[[ckn]] == 0, c++; s[[ckn]] = k]; n++]; n = 4; k++]; s]; seq[100] (* Amiram Eldar, Mar 06 2023 *)
|
|
CROSSREFS
|
Cf. A101321, A176774, A333914.
|
|
KEYWORD
|
nonn,new
|
|
AUTHOR
|
Ilya Gutkovskiy, Feb 15 2023
|
|
STATUS
|
approved
|
|
|
|
|
A360821
|
|
Number of primes of the form k^2+1 between n^2 and 2*n^2 exclusive.
|
|
+0
0
|
|
|
0, 1, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 3, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 3, 3, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,10
|
|
COMMENTS
|
This sequence studies the special case of Bertrand's postulate where n is a square and p of the form k^2+1.
Conjecture: Except for n = 7, between the square of an integer n > 1 and the double of its square there is always at least one prime number of the form k^2+1.
The following table shows the first 15 cases, with p prime of the form k^2+1.
+----+-------+-------+-----------------------------+------+
| n | n^2 | 2*n^2 | p such that n^2 < p < 2*n^2 | a(n) |
+----+-------+-------+-----------------------------+------+
| 1 | 1 | 2 | --------------------------- | 0 |
| 2 | 4 | 8 | 5 | 1 |
| 3 | 9 | 18 | 17 | 1 |
| 4 | 16 | 32 | 17 | 1 |
| 5 | 25 | 50 | 37 | 1 |
| 6 | 36 | 72 | 37 | 1 |
| 7 | 49 | 98 | --------------------------- | 0 |
| 8 | 64 | 128 | 101 | 1 |
| 9 | 81 | 162 | 101 | 1 |
| 10 | 100 | 200 | 101, 197 | 2 |
| 11 | 121 | 242 | 197 | 1 |
| 12 | 144 | 288 | 197, 257 | 2 |
| 13 | 169 | 338 | 197, 257 | 2 |
| 14 | 196 | 392 | 197, 257 | 2 |
| 15 | 225 | 450 | 257, 401 | 2 |
|
|
LINKS
|
Michel Lagneau, Table of n, a(n) for n = 1..1000
|
|
EXAMPLE
|
a(20) = 3 because there are 3 prime numbers of the form k^2+1 between 20^2 and 2*20^2. We obtain 400 < 401, 577, 677 < 800.
|
|
MAPLE
|
nn:=90:
for n
from 1 to nn do:
n1:=n^2:n2:=2*n^2:it:=0:
for p from n1+1 to n2-1 do:
p1:=sqrt(p-1):p2:=floor(p1):
if isprime(p) and p1=p2
then
it:=it+1:
else
fi:
od:
printf(`%d, `, it):
od:
|
|
PROG
|
(PARI) a(n) = my(nb=0); forprime(p=n^2+1, 2*n^2-1, if (issquare(p-1), nb++)); nb; \\ Michel Marcus, Feb 22 2023
|
|
CROSSREFS
|
Cf. A000290, A002496, A060715.
|
|
KEYWORD
|
nonn,new
|
|
AUTHOR
|
Michel Lagneau, Feb 21 2023
|
|
STATUS
|
approved
|
|
|
|
|
A361206
|
|
Lexicographically earliest infinite sequence of distinct imperfect numbers such that the sum of the abundance of all terms is never < 1.
|
|
+0
0
|
|
|
12, 1, 2, 4, 18, 3, 8, 20, 10, 24, 5, 7, 16, 30, 9, 14, 32, 36, 11, 13, 40, 15, 42, 17, 48, 19, 21, 54, 22, 44, 56, 50, 60, 23, 25, 52, 64, 66, 26, 70, 72, 27, 29, 34, 78, 45, 80, 33, 68, 84, 31, 35, 88, 90, 37, 38, 96, 39, 41, 100, 46, 102, 76, 104, 108, 43, 58
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
The abundance of each term is A033880(a(n)) and s = Sum_{i=1..n} A033880(a(i)).
All imperfect numbers A132999 will appear in this sequence and the abundant numbers A005101 will appear in order.
|
|
LINKS
|
Table of n, a(n) for n=1..67.
|
|
EXAMPLE
|
The sequence starts with a(1) = 12, since 12 is the first imperfect number with abundance greater than 0. Then the next term not yet in the sequence, such that s is not less than 1, is 1.
a(5) is the next abundant number 18, since any deficient number would bring s below 1.
n : 1 2 3 4 5 6 7 8 9 10
a(n): 12 1 2 4 18 3 8 20 10 24
s : 4 3 2 1 4 2 1 3 1 13
|
|
PROG
|
(Python)
from sympy.ntheory import abundance
from itertools import count, filterfalse
def A361206_list(nmax):
A, s = [], 0
for n in range(1, nmax+1):
A2 = set(A)
for y in filterfalse(A2.__contains__, count(1)):
ab = abundance(y)
if ab != 0 and ab + s >= 1:
A.append(y)
s += ab
break
return(A)
|
|
CROSSREFS
|
Cf. A005101, A033879, A033880, A132999.
|
|
KEYWORD
|
nonn,easy,new
|
|
AUTHOR
|
John Tyler Rascoe, Mar 04 2023
|
|
STATUS
|
approved
|
|
|
|
|
A361320
|
|
If n is composite, replace n with the concatenation of its nontrivial divisors, written in increasing order, each divisor being written in base 10 with its digits in reverse order, otherwise a(n) = n.
|
|
+0
0
|
|
|
1, 2, 3, 2, 5, 23, 7, 24, 3, 25, 11, 2346, 13, 27, 35, 248, 17, 2369, 19, 24501, 37, 211, 23, 2346821, 5, 231, 39, 24741, 29, 23560151, 31, 24861, 311, 271, 57, 234692181, 37, 291, 331, 24580102, 41, 23674112, 43, 241122, 35951, 232, 47, 23468216142, 7, 250152
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
First differs from A037279 at a(20).
|
|
LINKS
|
Table of n, a(n) for n=1..50.
|
|
EXAMPLE
|
Divisors of 20 are 1,2,4,5,10,20, so a(20)=24501.
|
|
PROG
|
(PARI) a(n) = if (isprime(n) || (n==1), return (n)); my(d=divisors(n), list=List()); for (i=2, #d-1, my(dd=digits(d[i])); forstep (j=#dd, 1, -1, listput(list, dd[j]))); fromdigits(Vec(list)); \\ Michel Marcus, Mar 09 2023
|
|
CROSSREFS
|
Cf. A037279, A130140.
|
|
KEYWORD
|
nonn,base,new
|
|
AUTHOR
|
Tyler Busby, Mar 09 2023
|
|
STATUS
|
approved
|
|
|
|
|
A361216
|
|
Triangle read by rows: T(n,k) is the maximum number of ways in which a set of integer-sided rectangular pieces can tile an n X k rectangle.
|
|
+0
0
|
|
|
1, 1, 4, 2, 11, 56, 3, 29, 370, 5752, 4, 94, 2666, 82310, 2519124, 6, 263, 19126, 1232770, 88117873, 6126859968, 12, 968, 134902, 19119198, 2835424200
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
Tilings that are rotations or reflections of each other are considered distinct.
Pieces can have any combination of integer side lengths, but for the optimal sets computed so far (up to (n,k) = (7,5)), all pieces have one side of length 1.
|
|
LINKS
|
Table of n, a(n) for n=1..26.
|
|
FORMULA
|
T(n,1) = A102462(n).
|
|
EXAMPLE
|
Triangle begins:
n\k| 1 2 3 4 5 6 7
---+-----------------------------------------------------
1 | 1
2 | 1 4
3 | 2 11 56
4 | 3 29 370 5752
5 | 4 94 2666 82310 2519124
6 | 6 263 19126 1232770 88117873 6126859968
7 | 12 968 134902 19119198 2835424200 ? ?
8 | 20 3416 1026667 307914196 109979838540 ? ?
A 3 X 3 square can be tiled by three 1 X 2 pieces and three 1 X 1 pieces in the following ways:
+---+---+---+ +---+---+---+ +---+---+---+
| | | | | | | | | | | |
+---+---+---+ + +---+---+ +---+ +---+
| | | | | | | | | | |
+---+---+ + +---+---+ + +---+---+ +
| | | | | | | | |
+---+---+---+ +---+---+---+ +---+---+---+
.
+---+---+---+ +---+---+---+ +---+---+---+
| | | | | | | | | |
+---+---+---+ +---+---+ + +---+---+---+
| | | | | | | | | |
+---+---+ + +---+---+---+ +---+---+---+
| | | | | | | | |
+---+---+---+ +---+---+---+ +---+---+---+
.
+---+---+---+ +---+---+---+
| | | | | |
+---+---+---+ +---+---+---+
| | | | | |
+---+---+---+ +---+---+---+
| | | | | |
+---+---+---+ +---+---+---+
The first six of these have no symmetries, so they account for 8 tilings each. The last two has a mirror symmetry, so they account for 4 tilings each. In total there are 6*8+2*4 = 56 tilings. This is the maximum for a 3 X 3 square, so T(3,3) = 56.
The following table shows the sets of pieces that give the maximum number of tilings up to (n,k) = (7,5). The solutions are unique except for (n,k) = (2,1) and (n,k) = (6,1).
\ Number of pieces of size
(n,k)\ 1 X 1 | 1 X 2 | 1 X 3 | 1 X 4
------+-------+-------+-------+------
(1,1) | 1 | 0 | 0 | 0
(2,1) | 2 | 0 | 0 | 0
(2,1) | 0 | 1 | 0 | 0
(2,2) | 2 | 1 | 0 | 0
(3,1) | 1 | 1 | 0 | 0
(3,2) | 2 | 2 | 0 | 0
(3,3) | 3 | 3 | 0 | 0
(4,1) | 2 | 1 | 0 | 0
(4,2) | 4 | 2 | 0 | 0
(4,3) | 3 | 3 | 1 | 0
(4,4) | 5 | 4 | 1 | 0
(5,1) | 3 | 1 | 0 | 0
(5,2) | 4 | 3 | 0 | 0
(5,3) | 4 | 4 | 1 | 0
(5,4) | 7 | 5 | 1 | 0
(5,5) | 7 | 6 | 2 | 0
(6,1) | 2 | 2 | 0 | 0
(6,1) | 1 | 1 | 1 | 0
(6,2) | 4 | 4 | 0 | 0
(6,3) | 7 | 4 | 1 | 0
(6,4) | 8 | 5 | 2 | 0
(6,5) | 10 | 7 | 2 | 0
(6,6) | 11 | 8 | 3 | 0
(7,1) | 2 | 1 | 1 | 0
(7,2) | 5 | 3 | 1 | 0
(7,3) | 8 | 5 | 1 | 0
(7,4) | 10 | 6 | 2 | 0
(7,5) | 11 | 7 | 2 | 1
|
|
CROSSREFS
|
Main diagonal: A361217.
Columns: A361218 (k = 2), A361219 (k = 3), A361220 (k = 4).
Cf. A360629, A361221.
|
|
KEYWORD
|
nonn,tabl,more,new
|
|
AUTHOR
|
Pontus von Brömssen, Mar 05 2023
|
|
STATUS
|
approved
|
|
|
|
|
A361360
|
|
Number of nonequivalent noncrossing caterpillars with n edges up to rotation and relection.
|
|
+0
0
|
|
|
1, 1, 1, 3, 7, 28, 104, 448, 1886, 8212, 35556, 155124, 675897, 2950074, 12872294, 56188904, 245253691, 1070581703, 4673231521, 20399699635, 89048927767, 388718917440, 1696845506274, 7407120344070, 32333775400516, 141144364258374, 616127577376396
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,4
|
|
COMMENTS
|
The number of all noncrossing caterpillars with n edges is given by A361356.
|
|
LINKS
|
Table of n, a(n) for n=0..26.
Index entries for linear recurrences with constant coefficients, signature (5,4,-34,7,63,-30,-46,31,13,-14,0,3,-1).
|
|
FORMULA
|
G.f.: (1 - 4*x - 8*x^2 + 28*x^3 + 15*x^4 - 55*x^5 - 2*x^6 + 46*x^7 - 11*x^8 - 19*x^9 + 10*x^10 + 2*x^11 - 2*x^12)/((1 - x)^2*(1 + x)^2*(1 - 5*x + 3*x^2 - x^3)*(1 - 5*x^2 + 3*x^4 - x^6)).
a(n) = 5*a(n-1) + 4*a(n-2) - 34*a(n-3) + 7*a(n-4) + 63*a(n-5) - 30*a(n-6) - 46*a(n-7) + 31*a(n-8) + 13*a(n-9) - 14*a(n-10) + 3*a(n-12) - a(n-13) for n >= 13.
|
|
PROG
|
(PARI)
G(x)={ my(f = x*(2 - x)/(1 - 5*x + 3*x^2 - x^3), g = 1 + x + x^2*(3 - 2*x + (4 - 3*x + x^2)*f + (1 + 2*x)*f^2)/(1 - x)^2); (intformal(g) - 3)/x/2 + x*subst((3 + 2*x*(3-x)*f)/(1-x)^2, x, x^2)/4 + subst(1/(1-x) + x*f/(1-x), x, x^2)/2}
{ Vec(G(x) + O(x^30)) }
|
|
CROSSREFS
|
Cf. A296533 (noncrossing trees), A361356, A361358, A361359 (up to rotation only).
|
|
KEYWORD
|
nonn,easy,new
|
|
AUTHOR
|
Andrew Howroyd, Mar 10 2023
|
|
STATUS
|
approved
|
|
|
|
|
A361359
|
|
Number of nonequivalent noncrossing caterpillars with n edges up to rotation.
|
|
+0
0
|
|
|
1, 1, 1, 4, 11, 49, 196, 868, 3721, 16306, 70891, 309739, 1350831, 5897934, 25740386, 112368153, 490489041, 2141121271, 9346382981, 40799215354, 178097506051, 777437032059, 3393689486976, 14814237183658, 64667544141561, 282288713218896, 1232255125682671
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,4
|
|
COMMENTS
|
The number of all noncrossing caterpillars with n edges is given by A361356.
|
|
LINKS
|
Table of n, a(n) for n=0..26.
Index entries for linear recurrences with constant coefficients, signature (6,-3,-26,36,-2,-18,6,5,-4,1).
|
|
FORMULA
|
G.f.: (1 - 5*x - 2*x^2 + 27*x^3 - 20*x^4 - 13*x^5 + 23*x^6 - 5*x^7 - 6*x^8 + 3*x^9)/((1 - x)*(1 - 5*x + 3*x^2 - x^3)*(1 - 5*x^2 + 3*x^4 - x^6)).
a(n) = 6*a(n-1) - 3*a(n-2) - 26*a(n-3) + 36*a(n-4) - 2*a(n-5) - 18*a(n-6) + 6*a(n-7) + 5*a(n-8) - 4*a(n-9) + a(n-10) for n >= 10.
|
|
PROG
|
(PARI)
G(x)={ my(f = x*(2 - x)/(1 - 5*x + 3*x^2 - x^3), g = 1 + x + x^2*(3 - 2*x + (4 - 3*x + x^2)*f + (1 + 2*x)*f^2)/(1 - x)^2); (intformal(g) - 3)/x + x*subst((1 + 2*x*f)/(1-x)^2, x, x^2)/2 }
{ Vec(G(x) + O(x^30)) }
|
|
CROSSREFS
|
Cf. A296532 (noncrossing trees), A361356, A361358, A361360 (up to rotation and reflection).
|
|
KEYWORD
|
nonn,easy,new
|
|
AUTHOR
|
Andrew Howroyd, Mar 09 2023
|
|
STATUS
|
approved
|
|
|
|
|
A361275
|
|
Number of 1423-avoiding even Grassmannian permutations of size n.
|
|
+0
0
|
|
|
1, 1, 1, 3, 5, 11, 17, 29, 41, 61, 81, 111, 141, 183, 225, 281, 337, 409, 481, 571, 661, 771, 881, 1013, 1145, 1301, 1457, 1639, 1821, 2031, 2241, 2481, 2721, 2993, 3265, 3571, 3877, 4219, 4561, 4941, 5321, 5741, 6161, 6623, 7085, 7591, 8097, 8649, 9201, 9801, 10401
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,4
|
|
COMMENTS
|
A permutation is said to be Grassmannian if it has at most one descent. A permutation is even if it has an even number of inversions.
Avoiding any of the patterns 2314 or 3412 gives the same sequence.
|
|
LINKS
|
Table of n, a(n) for n=0..50.
Juan B. Gil and Jessica A. Tomasko, Pattern-avoiding even and odd Grassmannian permutations, arXiv:2207.12617 [math.CO], 2022.
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).
|
|
FORMULA
|
G.f.: -(x^5-x^4-4*x^3+2*x^2+x-1)/((x+1)^2*(x-1)^4).
a(n) = 1 - 5*n/24 + n^3/12 - (-1)^n * n/8. - Robert Israel, Mar 10 2023
|
|
EXAMPLE
|
For n=4 the a(4) = 5 permutations are 1234, 1342, 2314, 3124, 3412.
|
|
MAPLE
|
seq(1 - 5*n/24 + n^3/12 - (-1)^n * n/8, n = 0 .. 100); # Robert Israel, Mar 10 2023
|
|
CROSSREFS
|
Cf. A356185, A361272, A361273, A361274.
For the corresponding odd permutations, cf. A005993.
|
|
KEYWORD
|
nonn,easy,new
|
|
AUTHOR
|
Juan B. Gil, Mar 10 2023
|
|
STATUS
|
approved
|
|
|
Search completed in 0.154 seconds
|