1,2

Deleting the least significant digit yields the (n-1)-st triangular number: a(n) = 5n(n-1)+1 = 10{n(n-1)/2} +1 = 10*T(n-1) +1. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), 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

Except for the first term, a(n)=10*n+a(n-1), (with a(1)=11) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Oct 24 2009]

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

Index entries for sequences related to centered polygonal numbers

5n(n-1)+1.

Binomial transform of [1, 10, 10, 0, 0, 0,...]; Narayana transform (A001263) of [1, 10, 0, 0, 0,...]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 29 2007

a(n)=10*n+a(n-1)-10 (with a(1)=1) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 10 2009]

For n=2, a(2)=10*2+1-10=11; n=3, a(3)=10*3+11-10=31; n=4, a(4)=10*4+31-10=61 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 10 2009]

lst={}; Do[p=(5*(n^2-n))+1; AppendTo[lst, p], {n, 5!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Sep 27 2008]

(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) ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Aug 11 2009]

Cf. A001263.

Sequence in context: A040162 A113747 A125239 this_sequence A090562 A136061 A090233

Adjacent sequences: A062783 A062784 A062785 this_sequence A062787 A062788 A062789

easy,nonn,new

Jason Earls (zevi_35711(AT)yahoo.com), Jul 19 2001

Better description from Terry Trotter, Apr 06, 2002.