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A231303
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Recurrence a(n) = a(n-2) + n^M for M=4, starting with a(0)=0, a(1)=1.
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10
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0, 1, 16, 82, 272, 707, 1568, 3108, 5664, 9669, 15664, 24310, 36400, 52871, 74816, 103496, 140352, 187017, 245328, 317338, 405328, 511819, 639584, 791660, 971360, 1182285, 1428336, 1713726, 2042992, 2421007, 2852992, 3344528, 3901568, 4530449, 5237904
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OFFSET
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0,3
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COMMENTS
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In physics, a(n)/2^(M-1) is the trace of the spin operator |S_z|^M for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and therefore the sum of their 4th powers is 2*82/16 = a(3)/8 (analogously for other values of M).
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LINKS
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FORMULA
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a(n) = Sum_{k=0..floor(n/2)} (n - 2*k)^4.
a(n) = (1/30)*n*(3*n^4 + 15*n^3 + 20*n^2 - 8).
G.f.: x*(1 + 10*x + x^2) / (1-x)^6.
(End)
E.g.f.: x*(30 + 210*x + 185*x^2 + 45*x^3 + 3*x^4)*exp(x)/30. - G. C. Greubel, Apr 24 2016
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EXAMPLE
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a(4) = 4^4 + 2^4 = 272; a(5) = 5^4 + 3^4 + 1^4 = 707.
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MATHEMATICA
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Table[SeriesCoefficient[x*(1+10*x+x^2)/(1-x)^6, {x, 0, n}], {n, 0, 40}] (* Michael De Vlieger, Dec 22 2015 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 16, 82, 272, 707}, 40] (* Vincenzo Librandi, Dec 23 2015 *)
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PROG
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(PARI) nmax=40; a = vector(nmax); a[2]=1; for(i=3, #a, a[i]=a[i-2]+(i-1)^4); print(a);
(PARI) concat(0, Vec(x*(1+10*x+x^2)/(1-x)^6 + O(x^40))) \\ Colin Barker, Dec 22 2015
(Magma) [1/30*n*(3*n^4+15*n^3+20*n^2-8): n in [0..40]]; // Vincenzo Librandi, Dec 23 2015
(Sage) [n*(3*n^4+15*n^3+20*n^2-8)/30 for n in (0..40)] # G. C. Greubel, Jul 01 2019
(GAP) List([0..40], n-> n*(3*n^4+15*n^3+20*n^2-8)/30) # G. C. Greubel, Jul 01 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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