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.

User: Daniel Carter

Daniel Carter's wiki page.

Daniel Carter has authored 6 sequences.

A350302 Irregular triangle read by rows: The n-th row lists the larger number in each "n-th power friendship", that is, numbers k such that digsum(digsum(k^n)^n) = k and digsum(k^n) < k.

Original entry on oeis.org

16, 28, 27, 29, 34, 47, 55, 62, 65, 73, 80, 127, 70, 121, 136, 181, 117, 130, 153, 153, 160, 161, 176, 181, 184, 187, 189, 198, 208, 221, 235, 193, 252, 190, 233, 220, 247, 220, 254, 257, 259, 277, 279, 289, 263, 319, 261, 331, 260, 297, 316, 280, 304, 313
Offset: 2

Author

Daniel Carter, Dec 23 2021

Keywords

Examples

			Triangle begins:
  16;
  28;
  27;
  29, 34;
  ;
  47, 55, 62, 65;
  73;
  80;
  ;
  ;
  127;
  70, 121, 136, 181;
  117, 130;
  153;
  153, 160;
  ...
		

Crossrefs

A350301 Irregular triangle read by rows: The n-th row lists the smaller number in each "n-th power friendship", that is, numbers k such that digsum(digsum(k^n)^n) = k and digsum(k^n) > k.

Original entry on oeis.org

13, 19, 18, 23, 31, 38, 44, 46, 56, 64, 35, 109, 52, 112, 118, 127, 97, 108, 144, 88, 144, 139, 152, 153, 154, 173, 178, 189, 172, 199, 203, 187, 234, 127, 188, 148, 238, 115, 229, 238, 239, 245, 234, 244, 245, 283, 252, 283, 161, 271, 288, 163, 259, 295, 316
Offset: 2

Author

Daniel Carter, Dec 23 2021

Keywords

Examples

			Triangle begins:
  13;
  19;
  18;
  23, 31;
  ;
  38, 44, 46, 56;
  64;
  35;
  ;
  ;
  109;
  52, 112, 118, 127;
  97, 108;
  144;
  88, 144;
  ...
		

Crossrefs

A350300 Irregular triangle read by rows: The n-th row lists the "n-th power friends", numbers k such that digsum(digsum(k^n)^n) = k but digsum(k^n) is not k.

Original entry on oeis.org

13, 16, 19, 28, 18, 27, 23, 29, 31, 34, 38, 44, 46, 47, 55, 56, 62, 65, 64, 73, 35, 80, 109, 127, 52, 70, 112, 118, 121, 127, 136, 181, 97, 108, 117, 130, 144, 153, 88, 144, 153, 160, 139, 152, 153, 154, 161, 173, 176, 178, 181, 184, 187, 189, 189, 198
Offset: 2

Author

Daniel Carter, Dec 23 2021

Keywords

Comments

Two numbers x and y with x < y are said to be "n-th power friends" if digsum(x^n)=y and digsum(y^n)=x. This sequence lists both x and y; A350301 lists just the x's and A350302 lists just the y's.
The name "n-th power friends" comes from "The Man Who Counted", where the n=2 case is discussed:
"The digits of the number 256 add up to 13. The square of 13 is 169. The digits of 169 add up to 16. As a result, the numbers 13 and 16 have a curious relation, which we could call a quadratic friendship" (p. 33).
There are finitely many such k for a particular n since digsum(digsum(k^n)^n) <= 9n log_10(9n log_10(k)). There are no such k for n=1 since either k is a single digit or else digsum(k) < k.

Examples

			Triangle begins:
  13, 16;
  19, 28;
  18, 27;
  23, 29, 31, 34;
  ;
  38, 44, 46, 47, 55, 56, 62, 65;
  64, 73;
  35, 80;
  ;
  ;
  109, 127;
  52, 70, 112, 118, 121, 127, 136, 181;
  97, 108, 117, 130;
  144, 153;
  88, 144, 153, 160;
  ...
18 and 27 are in row n=4 since 18^4 = 104976 and 1 + 0 + 4 + 9 + 7 + 6 = 27, and 27^4 = 531441 and 5 + 3 + 1 + 4 + 4 + 1 = 18.
		

References

  • M. Tahan, The Man Who Counted: A Collection of Mathematical Adventures, W. W. Norton & Company, 1993.

Crossrefs

Programs

  • Python
    from math import log
    n = 1
    while n <= 50:
        k = 2
        while 9*n*log(9*n*log(k,10),10) >= k:
            s1 = sum(int(d) for d in str(k**n))
            s2 = sum(int(d) for d in str(s1**n))
            if k != s1 and k == s2:
                print(k)
            k += 1
        n += 1

A344363 Decimal expansion of (5^(1/4) + 5^(3/4))/2.

Original entry on oeis.org

2, 4, 1, 9, 5, 2, 5, 1, 5, 3, 0, 5, 1, 6, 6, 5, 3, 3, 0, 9, 6, 4, 0, 3, 2, 1, 8, 0, 2, 1, 7, 0, 7, 6, 5, 3, 8, 6, 5, 1, 8, 1, 7, 8, 5, 7, 9, 3, 8, 5, 4, 7, 0, 8, 4, 6, 8, 3, 2, 5, 5, 3, 8, 2, 8, 9, 5, 8, 8, 4, 0, 4, 2, 5, 3, 9, 8, 9, 9, 6, 8, 5, 7, 3, 5, 8, 0, 1, 5, 5, 0, 8, 2, 4, 1, 8, 4, 6, 7, 8, 4, 8, 7, 3, 8
Offset: 1

Author

Daniel Carter, May 15 2021

Keywords

Comments

Solution for z in the system {x = 1/y + 1/z, y = x + 1/z, z = y + 1/x}. The corresponding values of x and y are (5^(1/4) + 5^(-1/4))/2 and 5^(1/4).
The largest aspect ratio of a set of three rectangles which have the property that any two of them can be scaled, rotated, and joined at an edge to obtain a rectangle with the third aspect ratio. The other two aspect ratios are given in the comment above.

Examples

			2.419525153051665330964032180217076538651...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[(Surd[5,4]+Surd[5^3,4])/2,10,120][[1]] (* Harvey P. Dale, Jan 01 2023 *)
  • PARI
    my(c=250+150*quadgen(20)); a_vector(len) = digits(sqrtint(floor(c*100^(len-2)))); \\ Kevin Ryde, May 28 2021

Formula

Equals sqrt(A090550).
Equals Gamma(1/20)*Gamma(9/20)/(Gamma(3/20)*Gamma(7/20)). [Zucker] - R. J. Mathar, Jun 24 2024

A344362 Decimal expansion of (5^(1/4) + 5^(-1/4))/2.

Original entry on oeis.org

1, 0, 8, 2, 0, 4, 4, 5, 4, 3, 0, 9, 8, 8, 2, 1, 2, 8, 2, 9, 5, 7, 5, 6, 6, 0, 3, 3, 6, 9, 9, 7, 8, 0, 6, 6, 5, 8, 7, 5, 7, 4, 7, 4, 7, 4, 6, 3, 3, 5, 9, 1, 9, 5, 5, 1, 4, 3, 2, 8, 8, 4, 7, 6, 5, 9, 8, 3, 4, 5, 3, 9, 5, 2, 9, 7, 1, 7, 8, 7, 4, 2, 2, 8, 6, 6, 1, 0, 0, 8, 0, 4, 4, 7, 8, 3, 2, 3, 2, 5, 9, 3, 0, 3, 4
Offset: 1

Author

Daniel Carter, May 15 2021

Keywords

Comments

Solution for x in the system {x = 1/y + 1/z, y = x + 1/z, z = y + 1/x}. The corresponding values of y and z are 5^(1/4) and (5^(1/4) + 5^(3/4))/2.
The smallest aspect ratio of a set of three rectangles which have the property that any two of them can be scaled, rotated, and joined at an edge to obtain a rectangle with the third aspect ratio. The other two aspect ratios are given in the comment above.

Examples

			1.082044543098821282957566033699780665875...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Cosh[Log[5]/4], 10, 120][[1]] (* Amiram Eldar, Jun 29 2023 *)
  • PARI
    solve(x=1, 2, 5*x^4 - 5*x^2 - 1) \\ Hugo Pfoertner, May 16 2021
    
  • PARI
    my(c=50+30*quadgen(20)); a_vector(len) = digits(sqrtint(floor(c*100^(len-2)))); \\ Kevin Ryde, May 28 2021

Formula

Equals 5*A293409.
Equals sqrt(A176015).
Equals cosh(log(5)/4). - Vaclav Kotesovec, May 28 2021

A340540 Number of walks of length 4n in the first octant using steps (1,1,1), (-1,0,0), (0,-1,0), and (0,0,-1) that start and end at the origin.

Original entry on oeis.org

1, 6, 288, 24444, 2738592, 361998432, 53414223552, 8525232846072, 1443209364298944, 255769050813120576, 47020653859202576640, 8907614785269428079168, 1730208409741026141405696, 343266632435192859791576064, 69350551439109880798294334208
Offset: 0

Author

Daniel Carter, Jan 10 2021

Keywords

Comments

There are no such walks with length that is not a multiple of 4.
a(n) is also the number of arrangements of n copies each of "a", "b", "c", and "d" such that no prefix has more b's, c's, or d's than a's.
The analogous problem in dimensions 1 and 2 are given respectively by A000108 (the Catalan numbers) and A006335.
No closed form is known. In fact, it is not known whether this sequence is D-finite (see Bacher et al.).

Crossrefs

Column k=3 of A340591.

Programs

  • Maple
    b:= proc(n, l) option remember; `if`(n=0, 1, `if`(add(i, i=l)+3 x+1, l)), 0) +add(`if`(l[i]>0,
          b(n-1, sort(subsop(i=l[i]-1, l))), 0), i=1..3))
        end:
    a:= n-> b(4*n, [0$3]):
    seq(a(n), n=0..15);  # Alois P. Heinz, Jan 12 2021
  • Mathematica
    b[n_, l_] := b[n, l] = If[n == 0, 1, If[Total[l] + 3 < n,
      b[n-1, l+1]], 0] + Sum[If[l[[i]] > 0,
      b[n-1, Sort[ReplacePart[l, i -> l[[i]]-1]]], 0], {i, 1, 3}] /. Null -> 0;
    a[n_] := b[4n, {0, 0, 0}];
    Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jul 10 2021, after Alois P. Heinz *)
  • Python
    import itertools as it
    i = 0
    while 1:
        counts = {(a,b,c):0 for a,b,c in it.product(range(i+1), repeat=3)}
        counts[0,0,0] = 1
        for _ in range(4*i):
            update = {(a,b,c):0 for a,b,c in it.product(range(i+1), repeat=3)}
            for x,y,z in counts:
                if counts[x,y,z] != 0:
                    for coord in [(x+1,y+1,z+1), (x-1,y,z), (x,y-1,z), (x,y,z-1)]:
                        if coord in update:
                            update[coord] += counts[x,y,z]
            counts = update
        print(i, counts[0,0,0])
        i += 1