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Partial sums of A101104. First differences of A005914.
6

%I #29 Jan 26 2022 02:29:42

%S 1,13,36,60,84,108,132,156,180,204,228,252,276,300,324,348,372,396,

%T 420,444,468,492,516,540,564,588,612,636,660,684,708,732,756,780,804,

%U 828,852,876,900,924,948,972,996,1020,1044,1068,1092,1116,1140,1164,1188,1212,1236,1260

%N Partial sums of A101104. First differences of A005914.

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

%C 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

%H G. C. Greubel, <a href="/A101103/b101103.txt">Table of n, a(n) for n = 1..5000</a>

%H Milan Janjic, <a href="https://pmf.unibl.org/wp-content/uploads/2017/10/enumfun.pdf">Enumerative Formulas for Some Functions on Finite Sets</a>.

%H Claudio de J. Pita Ruiz V., <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Pita/pita19.html">Some Number Arrays Related to Pascal and Lucas Triangles</a>, J. Int. Seq., Vol. 16 (2013) , Article 13.5.7.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).

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

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

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

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

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

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 157/156 - Pi/48. - _Amiram Eldar_, Jan 26 2022

%p 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

%t 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}]

%t Join[{1, 13},LinearRecurrence[{2, -1},{36, 60},33]] (* _Ray Chandler_, Sep 23 2015 *)

%o (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

%o (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

%o (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

%o (GAP) Concatenation([1,13],List([3..60],n->24*n-36)); # _Muniru A Asiru_, Dec 02 2018

%Y Cf. A073762.

%K easy,nonn

%O 1,2

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

%E Removed redundant information already in A101104. Reduced formulas by expansion of constants - _R. J. Mathar_, Nov 01 2010