A088058 Duplicate of A062392.
1, 15, 66, 190, 435, 861, 1540, 2556, 4005, 5995, 8646, 12090, 16471, 21945
Offset: 1
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a000583 = (^ 4) a000583_list = scanl (+) 0 a005917_list -- Reinhard Zumkeller, Nov 13 2014, Nov 11 2012
[n^4 : n in [0..50]]; // Wesley Ivan Hurt, Sep 05 2014
A000583 := n->n^4: seq(A000583(n), n=0..50); A000583:=-(z+1)*(z**2+10*z+1)/(z-1)**5; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero with (combinat):seq(fibonacci(3, n^2)-1, n=0..33); # Zerinvary Lajos, May 25 2008
Range[0,100]^4 (* Vladimir Joseph Stephan Orlovsky, Mar 14 2011 *)
makelist(n^4,n,0,30); /* Martin Ettl, Nov 12 2012 */
A000583(n) = n^4 \\ Michael B. Porter, Nov 09 2009
def a(n): return n**4 print([a(n) for n in range(34)]) # Michael S. Branicky, Nov 10 2022
a000584 = (^ 5) -- Reinhard Zumkeller, Nov 11 2012
Range[0,50]^5 (* Vladimir Joseph Stephan Orlovsky, Feb 20 2011 *)
makelist(n^5, n, 0, 30); /* Martin Ettl, Nov 12 2012 */
a(n)=n^5 \\ Charles R Greathouse IV, Jul 28 2015
[n**5 for n in range(30)] # Zerinvary Lajos, Jun 03 2009
From _Ilya Gutkovskiy_, Apr 13 2016: (Start) Illustration of initial terms: o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o n=0 n=1 n=2 n=3 n=4 (End) From _Klaus Purath_, Mar 18 2019: (Start) Examples: a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045). a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032). a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060). a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120). a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130). a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121). a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137). (End)
a028387 n = n + (n + 1) ^ 2 -- Reinhard Zumkeller, Jul 17 2014
[n + (n+1)^2: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011
FoldList[## + 2 &, 1, 2 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *) Table[n + (n + 1)^2, {n, 0, 100}] (* Vincenzo Librandi, Oct 17 2012 *) Table[ FrobeniusNumber[{n, n + 1}], {n, 2, 30}] (* Zak Seidov, Jan 14 2015 *)
a(n)=n^2+3*n+1 \\ Charles R Greathouse IV, Jun 10 2011
def a(n): return (n**2+3*n+1) # Torlach Rush, May 07 2024
[n+(n+1)^2 for n in range(0,48)] # Zerinvary Lajos, Jul 03 2008
1 x x^2 -1 x^3 -3x x^4 -6x^2 +5 x^5 -10x^3 +25x x^6 -15x^4 +75x^2 -61 x^7 -21x^5 +175x^3 -427x
w := proc(n,x) local v,k,pow,chen; pow := (a,b) -> if a = 0 and b = 0 then 1 else a^b fi; chen := proc(m) if irem(m+1,4) = 0 then RETURN(0) fi; 1/((-1)^iquo(m+1,4) *2^iquo(m,2)) end; add(add((-1)^v*binomial(k,v)*pow(v+x+1,n)*chen(k),v=0..k), k=0..n) end: # Coefficients with zeros: seq(print(seq(coeff(i!*coeff(series(exp(x*t)*sech(t),t,16),t,i),x,i-n),n=0..i)), i=0..8); # Recursion W := proc(n,z) option remember; local k,p; if n = 0 then 1 else p := irem(n+1,2); z^n - p + add(`if`(irem(k,2)=1,0, W(k,0)*binomial(n,k)*(power(z,n-k)-p)),k=2..n-1) fi end: # Peter Luschny, edited and additions Jul 07 2009, May 13 2010, Oct 24 2011
max = 9; rows = (Reverse[ CoefficientList[ #, x]] & ) /@ CoefficientList[ Series[ Exp[x*t]*Sech[t], {t, 0, max}], t]*Range[0, max]!; par[coefs_] := (p = Partition[ coefs, 2][[All, 1]]; If[ EvenQ[ Length[ coefs]], p, Append[ p, Last[ coefs]]]); Flatten[ par /@ rows] (* Jean-François Alcover, Oct 03 2011, after g.f. *) sk[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n-k), {k, 0, n}]; Table[CoefficientList[sk[n, x], x] // Reverse // Select[#, # =!= 0 &] &, {n, 0, 13}] // Flatten (* Jean-François Alcover, May 21 2013 *) Flatten@Table[Binomial[n, 2k] EulerE[2k], {n, 0, 12}, {k, 0, n/2}](* Oliver Seipel, Jan 14 2025 *)
def A046978(k): if k % 4 == 0: return 0 return (-1)**(k // 4) def A153641_poly(n, x): return expand(add(2**(-(k // 2))*A046978(k+1)*add((-1)**v*binomial(k,v)*(v+x+1)**n for v in (0..k)) for k in (0..n))) for n in (0..7): print(A153641_poly(n, x)) # Peter Luschny, Oct 24 2011
[n*(2*n^2 - 1): n in [0..40]]; // Vincenzo Librandi, Aug 18 2011
A007588:=n->n*(2*n^2 - 1); seq(A007588(n), n=0..40); # Wesley Ivan Hurt, Mar 10 2014
Table[ n(2n^2-1), {n,0,169} ] (* Alexander Adamchuk, Jun 02 2008 *) LinearRecurrence[{4,-6,4,-1},{0,1,14,51},50] (* Harvey P. Dale, Sep 16 2011 *)
a(n)=n*(2*n^2-1)
def A007588(n): return n*(2*n**2-1) # Chai Wah Wu, Feb 18 2022
[&+[n^9: n in [0..m]]: m in [0..22]]; // Bruno Berselli, Aug 23 2011
[seq(add(i^9,i=1..n),n=0..40)]; a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^9 od: seq(a[n], n=0..22); # Zerinvary Lajos, Feb 22 2008
lst={};s=0;Do[s=s+n^9;AppendTo[lst, s], {n, 10^2}];lst..or..Table[Sum[k^9, {k, 1, n}], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *) Accumulate[Range[0,30]^9] (* Harvey P. Dale, Oct 09 2016 *)
a(n)=n^2*(n+1)^2*(n^2+n-1)*(2*n^4+4*n^3-n^2-3*n+3)/20 \\ Charles R Greathouse IV, Oct 07 2015
A007487_list, m = [0], [362880, -1451520, 2328480, -1905120, 834120, -186480, 18150, -510, 1, 0, 0] for _ in range(10**2): for i in range(10): m[i+1]+= m[i] A007487_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
[n*(n+1)*(3*n^2+3*n-2)/8: n in [0..40]];
Table[n (n + 1) (3 n^2 + 3 n - 2)/8, {n, 0, 40}] LinearRecurrence[{5,-10,10,-5,1},{0,1,12,51,145},40] (* Harvey P. Dale, Aug 22 2016 *)
for(n=0, 40, print1(n*(n+1)*(3*n^2+3*n-2)/8", "));
[n*(n+1)*(n^4+2*n^3-2*n^2-3*n+3)/2: n in [0..50]]; // G. C. Greubel, Sep 01 2018
k=0;lst={k};Do[k=n^6-k;AppendTo[lst,k],{n,1,5!}];lst LinearRecurrence[{7,-21,35,-35,21,-7,1}, {0,1,63,666,3430,12195,34461}, 50] (* G. C. Greubel, Sep 01 2018 *) CoefficientList[Series[-((x (1+56 x+246 x^2+56 x^3+x^4))/(-1+x)^7),{x,0,30}],x] (* Harvey P. Dale, Aug 03 2024 *)
a(n)=n*(n+1)*(n^4+2*n^3-2*n^2-3*n+3)/2 \\ Charles R Greathouse IV, Oct 07 2015
a(4) = 4^4 + 2^4 = 272; a(5) = 5^4 + 3^4 + 1^4 = 707.
List([0..40], n-> n*(3*n^4+15*n^3+20*n^2-8)/30); # G. C. Greubel, Jul 01 2019
[1/30*n*(3*n^4+15*n^3+20*n^2-8): n in [0..40]]; // Vincenzo Librandi, Dec 23 2015
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 *)
nmax=40;a = vector(nmax);a[2]=1;for(i=3,#a,a[i]=a[i-2]+(i-1)^4); print(a);
concat(0, Vec(x*(1+10*x+x^2)/(1-x)^6 + O(x^40))) \\ Colin Barker, Dec 22 2015
[n*(3*n^4+15*n^3+20*n^2-8)/30 for n in (0..40)] # G. C. Greubel, Jul 01 2019
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