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.

Showing 1-3 of 3 results.

A234319 Smallest sum of n-th powers of k+1 consecutive positive integers that equals the sum of n-th powers of the next k consecutive integers, or -n if none.

Original entry on oeis.org

0, 3, 25, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, -13, -14, -15, -16, -17, -18, -19, -20, -21, -22, -23, -24, -25, -26, -27, -28, -29, -30, -31, -32, -33, -34, -35, -36, -37, -38, -39, -40, -41, -42, -43, -44, -45, -46, -47, -48, -49, -50, -51, -52, -53, -54
Offset: 0

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Author

Jonathan Sondow, Dec 23 2013

Keywords

Comments

a(n) is the smallest solution to m^n + (m+1)^n + ... + (m+k)^n = (m+k+1)^n + (m+k+2)^n + ... + (m+2*k)^n, or -n if no solution.
In 1879 Dostor gave all solutions for n = 2. In particular, a(2) = 25.
In 1906 Collignon proved that no solution exists for n = 3 and 4, so a(3) = -3 and a(4) = -4.
In 2013 Felten and Müller-Stach claimed to prove that no solution exists when n > 2, so if their proof is correct, a(n) = -n for n >= 3.

Examples

			m^0 + (m+1)^0 + ... + (m+k)^0 = k+1 > k = (m+k+1)^0 + (m+k+2)^0 + ... + (m+2*k)^0 for m > 0, so a(0) = -0 = 0.
1^1 + 2^1 = 3 = 3^1 is minimal for n = 1, so a(1) = 3.
3^2 + 4^2 = 25 = 5^2 is minimal for n = 2, so a(2) = 25.
		

References

  • Edouard Collignon, Note sur la résolution en entiers de m^2 + (m-r)^2 + ... + (m-kr)^2 = (m+r)^2 + ... + (m+kr)^2, Sphinx-Oedipe, 1 (1906-1907), 129-133.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[x*(27*x^3 - 50*x^2 + 19*x + 3)/(x - 1)^2, {x, 0, 50}], x] (* Wesley Ivan Hurt, Jun 21 2014 *)
  • PARI
    Vec(x*(27*x^3-50*x^2+19*x+3)/(x-1)^2 + O(x^100)) \\ Colin Barker, Apr 23 2014

Formula

a(0) = A059270(0) = A059255(0).
a(1) = A059270(1) = A230718(1).
a(2) = A059255(2) = A230718(2).
a(n) = -n for n > 2.
G.f.: x*(27*x^3-50*x^2+19*x+3) / (x-1)^2. - Colin Barker, Apr 23 2014

A344338 Smallest number that is the sum of two or more consecutive positive n-th powers in more than one way.

Original entry on oeis.org

9, 365, 33075
Offset: 1

Views

Author

Ilya Gutkovskiy, May 15 2021

Keywords

Comments

a(4) > 10^24. - Bert Dobbelaere, May 16 2021
Conjecture: no terms exist for n >= 4. - Jon E. Schoenfield, May 16 2021

Examples

			9 = 2 + 3 + 4 = 4 + 5.
365 = 10^2 + 11^2 + 12^2 = 13^2 + 14^2.
33075 = 11^3 + 12^3 + 13^3 + 14^3 + 15^3 + 16^3 + 17^3 + 18^3 + 19^3 = 15^3 + 16^3 + 17^3 + 18^3 + 19^3 + 20^3.
		

Crossrefs

Programs

  • Python
    N=3 # <== Adapt here
    import heapq
    sigma=1+2**N
    h=[(sigma,1,2)]
    nextcount=3
    oldv,olds,oldl=0,0,0
    while True:
        (v,s,l)=heapq.heappop(h)
        if v==oldv:
            break
        if v>=sigma:
            sigma += nextcount**N
            heapq.heappush(h, (sigma,1,nextcount))
            nextcount+=1
        oldv,olds,oldl = v,s,l
        v-=s**N ; s+=1 ; l+=1 ;    v+=l**N
        heapq.heappush(h,(v,s,l))
    print("a(%d) = %d = sum(i^%d, i=%d..%d) = sum(i^%d, i=%d..%d)"%
        (N,v,N,olds,oldl,N,s,l))
    # Bert Dobbelaere, May 16 2021

A350430 a(n) is the smallest n-th power which can be represented as the sum of n distinct positive n-th powers in exactly n ways, or -1 if none exists.

Original entry on oeis.org

1, 625, 157464
Offset: 1

Views

Author

Ilya Gutkovskiy, Dec 30 2021

Keywords

Comments

From Jon E. Schoenfield, Dec 30 2021: (Start)
222000^4 < a(4) < 4891341^4 = lcm(2829, 12259, 16359, 30381)^4 (see A039664, including the Wroblewski link).
10000^5 <= a(5) < 12528^5 = lcm(72, 1044, 1392, 2088, 3132)^5 (see A063923, including the Waldby link; note that, although the terms of A063923 include 72, 144, 1044, 1392, and 2088, whose LCM is only 4176, the primitive solution in which the sum of 5 distinct 5th powers is 144^5 is 0^5 + 27^5 + 84^5 + 110^5 + 133^5 = 144^5, which is not the sum of 5 positive n-th powers).
Conjecture: a(6) = -1. (End)

Examples

			For n = 2: 625 = 25^2 = 7^2 + 24^2 = 15^2 + 20^2.
For n = 3: 157464 = 54^3 = 6^3 + 36^3 + 48^3 = 12^3 + 19^3 + 53^3 = 27^3 + 36^3 + 45^3.
		

Crossrefs

Showing 1-3 of 3 results.