cp's OEIS Frontend

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.

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A118411 Numerator of sum of reciprocals of first n pentatope numbers A000332.

Original entry on oeis.org

1, 6, 19, 136, 83, 119, 656, 73, 190, 121, 1816, 559, 679, 815, 3872, 1139, 886, 513, 2360, 2023, 2299, 2599, 11696, 3275, 7306, 1353, 5992, 1653, 5455, 5983, 26176, 7139, 15538, 8435, 12184, 3293, 3553, 11479, 49360
Offset: 1

Views

Author

Jonathan Vos Post, Apr 27 2006

Keywords

Comments

Denominators are A118412. Fractions are: 1/1, 6/5, 19/15, 136/105, 83/63, 119/90, 656/495, 73/55, 190/143, 121/91, 1816/1365, 559/420, 679/510, 815/612, 3872/2907, 1139/855, 886/665, 513/385, 2360/1771, 2023/1518, 2299/1725, 2599/1950, 11696/8775, 3275/2457, 7306/5481, 1353/1015, 5992/4495, 1653/1240, 5455/4092, 5983/4488, 26176/19635, 7139/5355, 15538/11655, 8435/6327, 12184/9139, 3293/2470, 3553/2665, 11479/8610, 49360/37023. The denominator of sum of reciprocals of first n triangular numbers is A026741. The denominator of sum of reciprocals of first n tetrahedral numbers is A118392.

Examples

			a(1) = 1 = numerator of 1/1.
a(2) = 6 = numerator of 6/5 = 1/1 + 1/5.
a(3) = 19 = numerator of 19/15 = 1/1 + 1/5 + 1/15.
a(4) = 136 = numerator of 136/105 = 1/1 + 1/5 + 1/15 + 1/35.
a(5) = 55 = numerator of 55/42 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70.
a(10) = 190 = numerator of 190/143 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70 + 1/126 + 1/210 + 1/330 + 1/495 + 1/715.
a(20) = 2360 = numerator of 2360/1771 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70 + 1/126 + 1/210 + 1/330 + 1/495 + 1/715 + 1/1001 + 1/1365 + 1/1820 + 1/2380 + 1/3060 + 1/3876 + 1/4845 + 1/5985 + 1/7315 + 1/8855.

Crossrefs

Cf. A000332, A022998, A026741, A118391, A118391, A118412.

Programs

  • PARI
    s=0;for(i=4,50,s+=1/binomial(i,4);print(numerator(s))) /* Phil Carmody, Mar 27 2012 */

Formula

A118411(n)/A118412(n) = SUM[i=1..n] (1/A000332(n)). A118411(n)/A118412(n) = SUM[i=1..n] (1/C(n+2,4)). A118411(n)/A118412(n) = SUM[i=1..n] (1/(n*(n+1)*(n+2)*(n+3)/24)).

A118412 Denominator of sum of reciprocals of first n pentatope numbers A000332.

Original entry on oeis.org

1, 5, 15, 105, 42, 63, 90, 495, 55, 143, 91, 1365, 420, 510, 612, 2907, 855, 665, 385, 1771, 1518, 1725, 1950, 8775, 2457, 5481, 1015, 4495, 1240, 4092, 4488, 19635, 5355, 11655, 6327, 9139, 2470, 2665, 8610, 37023
Offset: 1

Views

Author

Jonathan Vos Post, Apr 27 2006

Keywords

Comments

Numerators are A118411. Fractions are: 1/1, 6/5, 19/15, 136/105, 83/63, 119/90, 656/495, 73/55, 190/143, 121/91, 1816/1365, 559/420, 679/510, 815/612, 3872/2907, 1139/855, 886/665, 513/385, 2360/1771, 2023/1518, 2299/1725, 2599/1950, 11696/8775, 3275/2457, 7306/5481, 1353/1015, 5992/4495, 1653/1240, 5455/4092, 5983/4488, 26176/19635, 7139/5355, 15538/11655, 8435/6327, 12184/9139, 3293/2470, 3553/2665, 11479/8610, 49360/37023. The denominator of sum of reciprocals of first n triangular numbers is A026741. The denominator of sum of reciprocals of first n tetrahedral numbers is A118392.

Examples

			a(1) = 1 = denominator of 1/1.
a(2) = 5 = denominator of 6/5 = 1/1 + 1/5.
a(3) = 15 = denominator of 19/15 = 1/1 + 1/5 + 1/15.
a(4) = 105 = denominator of 136/105 = 1/1 + 1/5 + 1/15 + 1/35.
a(5) = 42 = denominator of 55/42 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70.
a(10) = 143 = denominator of 190/143 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70 + 1/126 + 1/210 + 1/330 + 1/495 + 1/715.
a(20) = 1771 = denominator of 2360/1771 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70 + 1/126 + 1/210 + 1/330 + 1/495 + 1/715 + 1/1001 + 1/1365 + 1/1820 + 1/2380 + 1/3060 + 1/3876 + 1/4845 + 1/5985 + 1/7315 + 1/8855.

Crossrefs

Cf. A000332, A022998, A026741, A118391, A118391, A118411.

Programs

  • PARI
    s=0;for(i=4,50,s+=1/binomial(i,4);print(denominator(s))) /* Phil Carmody, Mar 27 2012 */

Formula

A118411(n)/A118412(n) = SUM[i=1..n] (1/A000332(n)). A118411(n)/A118412(n) = SUM[i=1..n] (1/C(n+2,4)). A118411(n)/A118412(n) = SUM[i=1..n] (1/(n*(n+1)*(n+2)*(n+3)/24)).

A283365 Minimal number of numbers in A000332 = { C(k,4); k=1,2,3,... } whose sum equals n.

Original entry on oeis.org

0, 1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 2, 3
Offset: 0

Views

Author

M. F. Hasler, Mar 06 2017

Keywords

Comments

Analog, for A000332 = {C(n,4)}, of A061336 (for triangular numbers A000217) and A104246 (for tetrahedral numbers A000292).

Links

Crossrefs

Cf. A000332 = {C(n,4)}; A061336 (analog for triangular numbers A000217), A104246 (analog for tetrahedral numbers A000292).

Programs

  • PARI
    {a(n,k=4,M=9e9,N=n) = (n <= k || M <= k+1) && return(n); for(m=k,M,binomial(m,k)>n && (M=m) && break); M-- <= k && return(n); my(b=binomial(M,k),c=binomial(M-1,k),NN); forstep( nn=n\b,0,-1, if(N>NN=nn+g(n-nn*b,k,M,N,d),N=NN); n-(nn-1)*b >= (N-nn+1)*c && break); N}

Formula

a(n) <= 8 = a(64) for all n, according to Kim (2003, first row of table "d = 4", p. 74), but this "numerical result" has no "* denoting exact values" (see Remark at end of paper), so it could be incorrect. [Disclaimer added by M. F. Hasler, Sep 22 2022]

A104397 Sums of 7 distinct positive pentatope numbers (A000332).

Original entry on oeis.org

462, 582, 666, 722, 747, 757, 777, 787, 791, 831, 887, 922, 942, 951, 952, 956, 967, 1007, 1042, 1051, 1062, 1072, 1076, 1091, 1107, 1126, 1142, 1146, 1156, 1160, 1162, 1171, 1172, 1176, 1182, 1202, 1212, 1216, 1227, 1237, 1247, 1251, 1253, 1262, 1267
Offset: 1

Views

Author

Jonathan Vos Post, Mar 05 2005

Keywords

Comments

Pentatope number Ptop(n) = binomial(n+3,4) = n*(n+1)*(n+2)*(n+3)/24.
Hyun Kwang Kim asserts that every positive integer can be represented as the sum of no more than 8 pentatope numbers; but in this sequence we are only concerned with sums of nonzero distinct pentatope numbers.

References

  • Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 55-57, 1996.

Links

Crossrefs

Cf. A000332, A100009, A102857, A104392, A104393, A104394, A104395, A104396.

Formula

a(n) = Ptop(e) + Ptop(f) + Ptop(g) + Ptop(h) + Ptop(i) + Ptop(j) + Ptop(k) for some positive e=/=f=/=g=/=h=/=i=/=j=/=k and Ptop(n) = binomial(n+3,4).

Extensions

Extended by Ray Chandler, Mar 05 2005

A192248 0-sequence of reduction of binomial coefficient sequence B(n,4)=A000332 by x^2 -> x+1.

Original entry on oeis.org

1, 1, 16, 51, 191, 569, 1619, 4259, 10694, 25709, 59743, 134818, 296798, 639518, 1352498, 2813750, 5769200, 11676395, 23358450, 46239770, 90667076, 176244326, 339887026, 650715076, 1237467151, 2338753519, 4394813644, 8214444389
Offset: 1

Views

Author

Clark Kimberling, Jun 27 2011

Keywords

Comments

See A192232 for definition of "k-sequence of reduction of [sequence] by [substitution]".

Crossrefs

Cf. A192232, A192249, A192069.

Programs

  • Mathematica
    c[n_] :=  n (n + 1) (n + 2) (n + 3)/24;  (* binomial B(n,4), A000332 *)
Table[c[n], {n, 1, 15}]
q[x_] := x + 1;
p[0, x_] := 1; p[n_, x_] := p[n - 1, x] + (x^n)*c[n + 1]
reductionRules = {x^y_?EvenQ -> q[x]^(y/2),
   x^y_?OddQ -> x q[x]^((y - 1)/2)};
t = Table[
  Last[Most[
    FixedPointList[Expand[#1 /. reductionRules] &, p[n, x]]]], {n, 0,
   40}]
Table[Coefficient[Part[t, n], x, 0], {n, 1, 40}]  (* A192248 *)
Table[Coefficient[Part[t, n], x, 1], {n, 1, 40}]  (* A192249 *)
Table[Coefficient[Part[t, n]/5, x, 1], {n, 1, 40}]  (* A192069 *)
(* by Peter J. C. Moses, Jun 20 2011 *)

Formula

Conjecture: G.f.: -x*(-1+5*x-20*x^2+30*x^3-25*x^4+8*x^5) / ( (x-1)*(x^2+x-1)^5 ). - R. J. Mathar, May 04 2014

A104398 Sums of 8 distinct positive pentatope numbers (A000332).

Original entry on oeis.org

792, 957, 1077, 1161, 1177, 1217, 1252, 1272, 1282, 1286, 1297, 1381, 1437, 1462, 1463, 1472, 1492, 1502, 1506, 1546, 1583, 1602, 1637, 1657, 1666, 1667, 1671, 1722, 1723, 1748, 1757, 1758, 1777, 1778, 1787, 1788, 1791, 1792, 1806, 1827, 1832, 1841
Offset: 1

Views

Author

Jonathan Vos Post, Mar 05 2005

Keywords

Comments

Pentatope number Ptop(n) = binomial(n+3,4) = n*(n+1)*(n+2)*(n+3)/24.
Hyun Kwang Kim asserts that every positive integer can be represented as the sum of no more than 8 pentatope numbers; but in this sequence we are only concerned with sums of nonzero distinct pentatope numbers.

References

  • Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 55-57, 1996.

Links

Crossrefs

Cf. A000332, A100009, A102857, A104392, A104393, A104394, A104395, A104396, A104397.

Programs

  • Mathematica
    Union[Total/@Subsets[Binomial[Range[4,15],4],{8}]] (* Harvey P. Dale, Mar 11 2012 *)

Formula

a(n) = Ptop(d) + Ptop(e) + Ptop(f) + Ptop(g) + Ptop(h) + Ptop(i) + Ptop(j) + Ptop(k) for some positive d=/=e=/=f=/=g=/=h=/=i=/=j=/=k and Ptop(n) = binomial(n+3,4).

Extensions

Extended by Ray Chandler, Mar 05 2005

A144865 Shadow transform of C(n+3,4) = A000332(n+3).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 4, 1, 2, 4, 4, 1, 4, 5, 6, 1, 4, 3, 4, 4, 5, 3, 4, 1, 4, 4, 2, 3, 4, 5, 4, 1, 5, 6, 16, 2, 4, 3, 6, 4, 4, 5, 4, 4, 9, 5, 4, 1, 4, 6, 7, 2, 4, 2, 16, 2, 7, 4, 4, 5, 4, 5, 11, 1, 16, 7, 4, 7, 6, 16, 4, 2, 4, 4, 5, 2, 16, 5, 4, 4, 2, 6, 4, 5, 16, 3, 5, 6, 4, 13, 16, 3, 6, 5, 16, 1, 4, 6, 10, 3, 4
Offset: 1

Views

Author

Alois P. Heinz, Sep 23 2008

Keywords

Links

  • Alois P. Heinz, Table of n, a(n) for n = 1..1000
  • Lorenz Halbeisen and Norbert Hungerbuehler, Number theoretic aspects of a combinatorial function, Notes on Number Theory and Discrete Mathematics 5(4) (1999), 138-150. (ps, pdf); see Definition 7 for the shadow transform.
  • N. J. A. Sloane, Transforms.

Crossrefs

4th column of A144871. Cf. A007318.

Programs

  • Maple
    shadow:= proc(p) proc(n) local j; add (`if` (modp(p(j), n)=0, 1,0), j=0..n-1) end end: f:= proc(k) proc(n) binomial (n+k-1,k) end end: a:= n-> shadow (f(4))(n): seq (a(n), n=1..120);

A192249 1-sequence of reduction of binomial coefficient sequence B(n,4)=A000332 by x^2 -> x+1.

Original entry on oeis.org

0, 5, 20, 90, 300, 930, 2610, 6900, 17295, 41605, 96660, 218145, 480225, 1034765, 2188385, 4552745, 9334760, 18892805, 37794765, 74817520, 146702410, 285169310, 549948760, 1052879110, 2002263910, 3784182685, 7110957850, 13291250220
Offset: 1

Views

Author

Clark Kimberling, Jun 27 2011

Keywords

Comments

See A192232 for definition of "k-sequence of reduction of [sequence] by [substitution]".

Crossrefs

Cf. A192232, A192248, A192069.

Programs

Formula

a(n) = 5*A192069(n).
Conjecture: G.f.: 5*x^2*(1-2*x+4*x^2-3*x^3+x^4) / ( (x-1)*(x^2+x-1)^5 ). - R. J. Mathar, May 04 2014

A104399 Sums of 9 distinct positive pentatope numbers (A000332).

Original entry on oeis.org

1287, 1507, 1672, 1792, 1793, 1876, 1932, 1958, 1967, 1987, 1997, 2001, 2078, 2157, 2162, 2178, 2218, 2253, 2273, 2283, 2287, 2298, 2322, 2382, 2438, 2442, 2463, 2473, 2493, 2503, 2507, 2526, 2542, 2547, 2582, 2603, 2612, 2617, 2637, 2638, 2647, 2651
Offset: 1

Views

Author

Jonathan Vos Post, Mar 05 2005

Keywords

Comments

Pentatope number Ptop(n) = binomial(n+3,4) = n*(n+1)*(n+2)*(n+3)/24. Hyun Kwang Kim asserts that every positive integer can be represented as the sum of no more than 8 pentatope numbers; but in this sequence we are only concerned with sums of nonzero distinct pentatope numbers.

References

  • Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 55-57, 1996.

Links

Crossrefs

Cf. A000332, A100009, A102857, A104392, A104393, A104394, A104395, A104396, A104397, A104398.

Formula

a(n) = Ptop(c) + Ptop(d) + Ptop(e) + Ptop(f) + Ptop(g) + Ptop(h) + Ptop(i) + Ptop(j) + Ptop(k) for some positive c=/=d=/=e=/=f=/=g=/=h=/=i=/=j=/=k and Ptop(n) = binomial(n+3,4).

Extensions

Extended by Ray Chandler, Mar 05 2005

A104400 Sums of 10 distinct positive pentatope numbers (A000332).

Original entry on oeis.org

2002, 2288, 2508, 2652, 2673, 2793, 2872, 2877, 2933, 2968, 2988, 2998, 3002, 3037, 3107, 3157, 3158, 3241, 3297, 3323, 3327, 3332, 3352, 3362, 3366, 3443, 3492, 3527, 3543, 3583, 3612, 3613, 3618, 3638, 3648, 3652, 3663, 3667, 3696, 3747, 3752, 3778
Offset: 1

Views

Author

Jonathan Vos Post, Mar 05 2005

Keywords

Comments

Pentatope number Ptop(n) = binomial(n+3,4) = n*(n+1)*(n+2)*(n+3)/24.
Hyun Kwang Kim asserts that every positive integer can be represented as the sum of no more than 8 pentatope numbers, but in this sequence we are only concerned with sums of nonzero distinct pentatope numbers.

References

  • Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 55-57, 1996.

Links

Crossrefs

Cf. A000332, A100009, A102857, A104392, A104393, A104394, A104395, A104396, A104397, A104398, A104399.

Programs

  • Maple
    N:= 10000: # to get all terms <= N
nmax:= floor(-3/2+1/2*sqrt(5+4*sqrt(1+24*N))):
S:= select(`<=`,{seq(add(s*(s+1)*(s+2)*(s+3)/24,s=c),
     c = combinat:-choose(nmax,10))},N):
sort(convert(S,list)); # Robert Israel, Dec 14 2015

Formula

a(n) = Ptop(b) + Ptop(c) + Ptop(d) + Ptop(e) + Ptop(f) + Ptop(g) + Ptop(h) + Ptop(i) + Ptop(j) + Ptop(k) for some positive b=/=c=/=d=/=e=/=f=/=g=/=h=/=i=/=j=/=k and Ptop(n) = binomial(n+3,4).

Extensions

Extended by Ray Chandler, Mar 05 2005
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