1,2

Deleting the least significant digit yields the (n-1)-st triangular number: a(n) = 10*A000217(n-1) + 1. - Amarnath Murthy, Dec 11 2003

All divisors of a(n) are congruent to 1 or -1, modulo 10; that is, they end in the decimal digit 1 or 9. Proof: If p is an odd prime different from 5 then 5n^2 - 5n + 1 == 0 (mod p) implies 25(2n - 1)^2 == 5 (mod p), whence p == 1 or -1 (mod 10). - Nick Hobson, Nov 13 2006

Centered decagonal numbers. - Omar E. Pol, Oct 03 2011

The partial sums of this sequence give A004466. - Leo Tavares, Oct 04 2021

The continued fraction expansion of sqrt(5*a(n)) is [5n-3; {2, 2n-2, 2, 10n-6}]. For n=1, this collapses to [2; {4}]. - Magus K. Chu, Sep 12 2022

T. D. Noe, Table of n, a(n) for n = 1..1000

Leo Tavares, Illustration: Pentagonal Stars

Leo Tavares, Illustration: Mid-section Stars

Leo Tavares, Illustration: Mid-section Star Pillars

Leo Tavares, Illustration: Trapezoidal Rays

Index entries for sequences related to centered polygonal numbers

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

a(n) = 5*n*(n-1) + 1.

From Gary W. Adamson, Dec 29 2007: (Start)

Binomial transform of [1, 10, 10, 0, 0, 0, ...];

Narayana transform (A001263) of [1, 10, 0, 0, 0, ...]. (End)

a(n) = 10*n + a(n-1) - 10; a(1)=1. - Vincenzo Librandi, Aug 07 2010

G.f.: x*(1+8*x+x^2) / (1-x)^3. - R. J. Mathar, Feb 04 2011

a(n) = A124080(n-1) + 1. - Omar E. Pol, Oct 03 2011

a(n) = A101321(10,n-1). - R. J. Mathar, Jul 28 2016

a(n) = A028387(A016861(n-1))/5 for n > 0. - Art Baker, Mar 28 2019

E.g.f.: (1+5*x^2)*exp(x) -1. - G. C. Greubel, Mar 30 2019

Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(5))) / sqrt(5). - Vaclav Kotesovec, Jul 23 2019

From Amiram Eldar, Jun 20 2020: (Start)

Sum_{n>=1} a(n)/n! = 6*e - 1.

Sum_{n>=1} (-1)^n * a(n)/n! = 6/e - 1. (End)

a(n) = A005891(n-1) + 5*A000217(n-1). - Leo Tavares, Jul 14 2021

a(n) = A003154(n) - 2*A000217(n-1). See Mid-section Stars illustration. - Leo Tavares, Sep 06 2021

From Leo Tavares, Oct 06 2021: (Start)

a(n) = A144390(n-1) + 2*A028387(n-1). See Mid-section Star Pillars illustration.

a(n) = A000326(n) + A000217(n) + 3*A000217(n-1). See Trapezoidal Rays illustration.

a(n) = A060544(n) + A000217(n-1). (End)

From Leo Tavares, Oct 31 2021: (Start)

a(n) = A016754(n-1) + 2*A000217(n-1)

a(n) = A016754(n-1) + A002378(n-1)

a(n) = A069099(n) + 3*A000217(n-1)

a(n) = A069099(n) + A045943(n-1)

a(n) = A003215(n-1) + 4*A000217(n-1)

a(n) = A003215(n-1) + A046092(n-1)

a(n) = A001844(n-1) + 6*A000217(n-1)

a(n) = A001844(n-1) + A028896(n-1)

a(n) = A005448(n) + 7*A000217(n)

a(n) = A005448(n) + A024966(n). (End)

FoldList[#1+#2 &, 1, 10Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)

1+5*Pochhammer[Range[50]-1, 2] (* G. C. Greubel, Mar 30 2019 *)

(PARI) j=[]; for(n=1, 75, j=concat(j, (5*n*(n-1)+1))); j

(PARI) for (n=1, 1000, write("b062786.txt", n, " ", 5*n*(n - 1) + 1) ) \\ Harry J. Smith, Aug 11 2009

(Magma) [1+5*n*(n-1): n in [1..50]]; // G. C. Greubel, Mar 30 2019

(Sage) [1+5*rising_factorial(n-1, 2) for n in (1..50)] # G. C. Greubel, Mar 30 2019

(GAP) List([1..50], n-> 1+5*n*(n-1)) # G. C. Greubel, Mar 30 2019

Cf. A001263.

Cf. A124080, A101321.

Cf. A028387, A016861.

Cf. A003154, A005891, A000217.

Cf. A004466.

Cf. A144390, A028387.

Cf. A000326.

Cf. A060544.

Cf. A016754, A002378, A069099, A045943, A003215, A046092, A001844, A028896, A005448, A024966.

Sequence in context: A113747 A202007 A125239 * A090562 A174244 A136061

Adjacent sequences: A062783 A062784 A062785 * A062787 A062788 A062789

easy,nonn

Jason Earls, Jul 19 2001

Better description from Terrel Trotter, Jr., Apr 06 2002

approved