A101103 Partial sums of A101104. First differences of A005914.

1, 13, 36, 60, 84, 108, 132, 156, 180, 204, 228, 252, 276, 300, 324, 348, 372, 396, 420, 444, 468, 492, 516, 540, 564, 588, 612, 636, 660, 684, 708, 732, 756, 780, 804, 828, 852, 876, 900, 924, 948, 972, 996, 1020, 1044, 1068, 1092, 1116, 1140, 1164, 1188, 1212, 1236, 1260

Offset: 1

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Keywords

• easy
• nonn

Comments

For more information, cross-references etc., see A101104.

For n >= 3, a(n) is equal to the number of functions f:{1,2,3,4}->{1,2,...,n} such that Im(f) contains 3 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Mar 08 2007

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000
• Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
• Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq., Vol. 16 (2013) , Article 13.5.7.
• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Crossrefs

Cf. A073762.

Programs

GAP

Concatenation([1,13],List([3..60],n->24*n-36)); # Muniru A Asiru, Dec 02 2018

Magma

I:=[36,60]; [1,13] cat [n le 2 select I[n] else 2*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 01 2018

Maple

seq(coeff(series(x*(1+x)*(1+10*x+x^2)/(1-x)^2,x,n+1), x, n), n = 1 .. 60); # Muniru A Asiru, Dec 02 2018

Mathematica

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 2, 2}, {k, 0, 34}] OR SeriesAtLevelR = Sum[Eulerian[n, i - 1]*Binomial[n + x - i + r, n + r], {i, 1, n}]; Table[SeriesAtLevelR, {n, 4, 4}, {r, -3, -3}, {x, 3, 35}]
Join[{1, 13},LinearRecurrence[{2, -1},{36, 60},33]] (* Ray Chandler, Sep 23 2015 *)

PARI

my(x='x+O('x^60)); Vec(x*(1+x)*(1+10*x+x^2)/(1-x)^2) \\ G. C. Greubel, Dec 01 2018

Sage

s=(x*(1+x)*(1+10*x+x^2)/(1-x)^2).series(x, 30); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 01 2018

Formula

a(n) = 2*a(n-1) - a(n-2), n > 4.

G.f.: x*(1+x)*(1 + 10*x + x^2)/(1-x)^2.

a(n) = 24*n - 36, n >= 3.

a(n) = Sum_{j=0..n} (-1)^j*binomial(3, j)*(n - j)^4. [Indices shifted, Nov 01 2010]

a(n) = Sum_{i=1..4} A008292(4,i)*binomial(n-i+1,1). [Indices shifted, Nov 01 2010]

Sum_{n>=1} (-1)^(n+1)/a(n) = 157/156 - Pi/48. - Amiram Eldar, Jan 26 2022

Extensions

Removed redundant information already in A101104. Reduced formulas by expansion of constants - R. J. Mathar, Nov 01 2010