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A030662
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Number of combinations of n things from 1 to n at a time, with repeats allowed.
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22
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1, 5, 19, 69, 251, 923, 3431, 12869, 48619, 184755, 705431, 2704155, 10400599, 40116599, 155117519, 601080389, 2333606219, 9075135299, 35345263799, 137846528819, 538257874439, 2104098963719, 8233430727599, 32247603683099, 126410606437751, 495918532948103
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OFFSET
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1,2
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COMMENTS
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Add terms of an increasingly bigger diamond-shaped part of Pascal's triangle:
.......................... 1
............ 1 .......... 1 1
.. 1 ...... 1 1 ........ 1 2 1
. 1 1 =5 . 1 2 1 =19 .. 1 3 3 1 =69
.. 2 ...... 3 3 ........ 4 6 4
............ 6 ......... 10 10
.......................... 20
- Ralf Stephan, May 17 2004
The prime p divides a((p-1)/2) for p = 5,13,17,29,37,41,53,61,73,89,97.. = A002144[n] Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006
Also, number of square submatrices of a square matrix. - Jono Henshaw (jjono(AT)hotmail.com), Apr 22 2008
Partial sums of A051924. - J. M. Bergot, Jun 22 2013
Number of partitions with Ferrers diagrams that fit in an n X n box (
excluding the empty partition of 0). - Michael Somos, Jun 02 2014
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LINKS
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T. D. Noe, Table of n, a(n) for n = 1..500
J. D. Horton and A. Kurn, Counting sequences with complete increasing subsequences, Congress Numerantium, 33 (1981), 75-80. MR 681905
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From N. J. A. Sloane, Feb 13 2013
M. Janjic, B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5
Jianqiang Zhao, Uniform Approach to Double Shuffle and Duality Relations of Various q-Analogs of Multiple Zeta Values via Rota-Baxter Algebras, arXiv preprint arXiv:1412.8044 [math.NT], 2014.
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FORMULA
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a(n) = A000984(n) - 1.
a(n) = 2*(2*n-1)!/(n!*(n-1)!)-1.
a(n) = Sum_{k=1..n} binomial(n, k)^2. - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{j=0..n} Sum_{i=j..n+j} binomial(i, j). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jul 23 2003
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-1} binomial(i+j, i). - N. J. A. Sloane, Jan 31 2009
Also for n>1: a(n)=(2*n)!/(n!)^2-1 - Hugo Pfoertner, Feb 10 2004
a(n) = Sum[Sum[(2n-i-j)!/(n-i)!/(n-j)!,{i,1,n}],{j,1,n}]. - Alexander Adamchuk, Jul 04 2006
a(n) = A115112(n) + 1. - Jono Henshaw (jjono(AT)hotmail.com), Apr 22 2008
G.f.: Q(0)*(1-4*x)/x - 1/x/(1-x), where Q(k)= 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
Conjecture: n*a(n) +2*(-3*n+2)*a(n-1) +(9*n-14)*a(n-2) +2*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Jun 25 2013
0 = a(n)*(+16*a(n+1) - 70*a(n+2) + 68*a(n+3) - 14*a(n+4)) + a(n+1)*(-2*a(n+1) + 61*a(n+2) - 96*a(n+3) + 23*a(n+4)) + a(n+2)*(-6*a(n+2) + 31*a(n+3) - 10*a(n+4)) + a(n+3)*(-2*a(n+3) + a(n+4)) for all n in Z. - Michael Somos, Jun 02 2014
From Ilya Gutkovskiy, Jan 25 2017: (Start)
O.g.f.: (1 - x - sqrt(1 - 4*x))/((1 - x)*sqrt(1 - 4*x)).
E.g.f.: exp(x)*(exp(x)*BesselI(0,2*x) - 1). (End)
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EXAMPLE
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G.f. = x + 5*x^2 + 19*x^3 + 69*x^4 + 251*x^5 + 923*x^6 + 3431*x^7 + ...
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MAPLE
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seq(sum((binomial(n, m))^2, m=1..n), n=1..23); # Zerinvary Lajos, Jun 19 2008
f:=n->add( add( binomial(i+j, i), i=0..n), j=0..n); [seq(f(n), n=0..12)]; # N. J. A. Sloane, Jan 31 2009
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MATHEMATICA
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Table[Sum[Sum[(2n-i-j)!/(n-i)!/(n-j)!, {i, 1, n}], {j, 1, n}], {n, 1, 20}] (* Alexander Adamchuk, Jul 04 2006 *)
a[n_] := 2*(2*n-1)!/(n*(n-1)!^2)-1; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Oct 11 2012, from first formula *)
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PROG
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(Sage)
def a(n) : return binomial(2*n, n) - 1
[a(n) for n in (1..26)] # Peter Luschny, Apr 21 2012
(PARI) a(n)=binomial(2*n, n)-1 \\ Charles R Greathouse IV, Jun 26 2013
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CROSSREFS
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2*A001700 - 1.
Column k=2 of A047909.
Cf. A091908, A144660, A002144.
Central column of triangle A014473.
Right-hand column 2 of triangle A102541.
Sequence in context: A240525 A264200 A055991 * A149758 A026590 A243413
Adjacent sequences: A030659 A030660 A030661 * A030663 A030664 A030665
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KEYWORD
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nonn,nice
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AUTHOR
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Donald Mintz (djmintz(AT)home.com)
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STATUS
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approved
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