Lucas Colucci - cp's OEIS frontend

A335824 Persistence of the 1-shifted Sloane's problem: number of iterations of "multiply together all the digits of a number (in base 10) shifted by +1" needed to reach a fixed point or a cycle.

2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 0, 2, 1, 1, 1, 2, 3, 1, 2, 4, 5, 2, 1, 1, 2, 3, 2, 4, 6, 3, 7, 2, 1, 1, 3, 2, 2, 2, 5, 2, 3, 2, 1, 2, 1, 4, 2, 7, 4, 4, 3, 2, 1, 2, 2, 6, 5, 4, 3, 5, 7, 2, 1, 3, 4, 3, 2, 4, 5, 6, 5, 2, 1, 1, 5, 7, 3, 3, 7, 5, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

Lucas Colucci, Jun 25 2020

The sequence can also be defined as the number of iterations of A089898 required to reach a fixed point or a cycle.
Wagstaff proved that a(n) is well-defined for every n; i.e., every number eventually converges to a fixed point or a cycle when iterating its digits shifted by 1. Moreover, the only fixed point is 18 and the only cycle is (2,3,...,10).
It is likely, but not known, that this sequence is unbounded.

			17->16->14->10, which belongs to the cycle (2,3,...,10). Thus, a(17)=3.
44->25->18, which is a fixed point. Thus, a(44)=2.

Gabriel Bonuccelli, Lucas Colucci, and Edson de Faria, On the Erdős-Sloane and Shifted Sloane Persistence, arXiv:2009.01114 [math.NT], 2020.
Samuel S. Wagstaff, Iterating the product of shifted digits, Fibonacci Quarterly 19.4 (1981): 340-347.

Cf. A089898.

g:= n -> convert(map(`+`,convert(n,base,10),1),`*`):
f:= proc(n)
  local k, x, R;
  x:= n;
  R[x]:= 0;
  for k from 1 do
    x:= g(x);
    if assigned(R[x]) then return R[x] fi;
    R[x]:= k;
  od;
end proc:
map(f, [$0..100]); # Robert Israel, Jun 25 2020

A335808 Nonzero multiplicative persistence in base 10: number of iterations of "multiply nonzero digits in base 10" needed to reach a number < 10.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 1, 1, 1, 2, 2, 2, 2, 3, 2, 3, 1, 1, 2, 2, 2, 3, 2, 3, 2, 3, 1, 1, 2, 2, 2, 2, 3, 2, 3, 3, 1, 1, 2, 2, 3, 3, 2, 4, 3, 3, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 1, 1, 2, 3, 3, 3, 3, 3, 3, 2

Offset: 0

Lucas Colucci, Jun 24 2020

Coincides with A031346 up to n=204.

Differs from A087472 first at n=110. - R. J. Mathar, Aug 10 2020

R. K. Guy, Unsolved Problems in Number Theory, E16, pages 262-263.

Lucas Colucci, Table of n, a(n) for n = 0..9999
Gabriel Bonuccelli, Lucas Colucci, and Edson de Faria, On the Erdős-Sloane and Shifted Sloane Persistence, arXiv:2009.01114 [math.NT], 2020.

Cf. A031346, A051801.

A335808 := proc(n)
    option remember;
    if n < 10 then
        0 ;
    elif n < 20 then
        1 ;
    else
        A051801(n) ;
        1+procname(%) ;
    end if;
end proc: # R. J. Mathar, Aug 10 2020

Array[-1 + Length[NestWhileList[Times @@ DeleteCases[IntegerDigits[#], ?(# == 0 &)] &, #, # >= 10 &]] &, 105, 0] (* _Michael De Vlieger, Jun 24 2020 *)

a(n) = for (k=0, oo, if (n<10, return (k), n=vecprod(select(sign, digits(n))))) \\ Rémy Sigrist, Jul 18 2020

A106793 Number of words (over an alphabet of size 26) of length n with all different letters.

Original entry on oeis.org

1, 26, 650, 15600, 358800, 7893600, 165765600, 3315312000, 62990928000, 1133836704000, 19275223968000, 308403583488000, 4626053752320000, 64764752532480000, 841941782922240000, 10103301395066880000

Offset: 0

Luca Colucci, May 17 2005

			a(2) = 650 because 26! / 24! = 650

Cf. A106710.

a(n) = 26! / (26 - n)!

A108297 Least positive k such that k * n^2 ends with n, or 0 no such k exists.

Original entry on oeis.org

1, 1, 3, 7, 4, 1, 1, 3, 2, 9, 0, 91, 23, 77, 0, 0, 11, 53, 0, 79, 0, 81, 0, 87, 24, 1, 0, 63, 17, 69, 0, 71, 18, 97, 0, 0, 16, 73, 0, 59, 0, 61, 0, 7, 4, 0, 0, 83, 12, 49, 0, 51, 13, 17, 0, 0, 21, 93, 0, 39, 0, 41, 0, 27, 9, 0, 0, 3, 7, 29, 0, 31, 8, 37, 0, 3, 1, 13, 0, 19, 0, 21, 0, 47, 14, 0

Offset: 0

Luca Colucci, Jun 29 2005

			a(3) = 7 because 7 * 3^2 = 63, which ends with "3".
a(11) = 91 because 91 * 11^2 = 11011 which ends with "11".
a(22) = 0 because there is no "k" such that k*22^2 ends with 22.

f[n_] := Block[{k = 1, n2 = n^2, m = Max[10^Floor[ Log[10, n] + 1], 10]}, While[k < 10^4 && Mod[k*n2, m] != n, k++ ]; If[k == 10^4, 0, k]]; Table[ f[n], {n, 85}] (* Robert G. Wilson v, Jul 02 2005 *)

Edited and extended by Robert G. Wilson v, Jul 02 2005

A108343 Gapful numbers >= 100: numbers that are divisible by the number formed by their first and last digit. Numbers up to 100 trivially have this property and are excluded.

Original entry on oeis.org

100, 105, 108, 110, 120, 121, 130, 132, 135, 140, 143, 150, 154, 160, 165, 170, 176, 180, 187, 190, 192, 195, 198, 200, 220, 225, 231, 240, 242, 253, 260, 264, 275, 280, 286, 297, 300, 315, 330, 341, 352, 360, 363, 374, 385, 390, 396, 400, 405, 440, 451

Offset: 1

Luca Colucci, Jul 01 2005

			253 is in the sequence because 253 = 11 * 23 and 23 is the concatenation of 2 and 3 (first and last digit of 253).

Aaron Toponce, Table of n, a(n) for n = 1..10000

fQ[ n_ ] := Block[ {id = IntegerDigits[ n ]}, IntegerQ[ n / FromDigits[ {id[ [ 1 ] ], id[ [ -1 ] ]} ] ] ]; Select[ Range[ 100, 461 ], fQ[ # ] & ] (* Robert G. Wilson v, Jul 19 2005 *)

Corrected and extended by Robert G. Wilson v, Jul 19 2005

A106710 Number of words with n letters from an alphabet of size 26 with at least two equal consecutive letters.

Original entry on oeis.org

0, 26, 1326, 50726, 1725126, 55009526, 1684153926, 50135658326, 1462218522726, 41984966747126, 1190791264331526, 33440126095275926, 931432109043580326, 25766955599293244726, 708683864685628269126, 19394355959426432653526, 528467641885089690397926

Offset: 1

Luca Colucci, May 14 2005

			a(3) = 1326 because 26^3 - 26*(25^2) = 1326.

Colin Barker, Table of n, a(n) for n = 1..706
Index entries for linear recurrences with constant coefficients, signature (51,-650).

Cf. A009969, A009970,

Table[26*(26^(n-1) -25^(n-1)), {n, 25}] (* G. C. Greubel, Sep 10 2021 *)

a(n) = 26^n - 26*(25^(n - 1)); \\ Michel Marcus, Aug 14 2013

concat(0, Vec(26*x^2/((25*x-1)*(26*x-1)) + O(x^100))) \\ Colin Barker, Nov 05 2015

[26*(26^(n-1) - 25^(n-1)) for n in (1..25)] # G. C. Greubel, Sep 10 2021

a(n) = 26^n - 26*25^(n - 1).
From Colin Barker, Nov 05 2015: (Start)
a(n) = 51*a(n-1) - 650*a(n-2) for n>2.
G.f.: 26*x^2 / ((1-25*x)*(1-26*x)). (End)
From G. C. Greubel, Sep 10 2021: (Start)
a(n) = 26*(A009970(n-1) - A009969(n-1)).
E.g.f.: exp(26*x) - (26/25)*exp(25*x). (End)

More terms from Michel Marcus, Aug 14 2013

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User: Lucas Colucci

A335824 Persistence of the 1-shifted Sloane's problem: number of iterations of "multiply together all the digits of a number (in base 10) shifted by +1" needed to reach a fixed point or a cycle.

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A335808 Nonzero multiplicative persistence in base 10: number of iterations of "multiply nonzero digits in base 10" needed to reach a number < 10.

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A106793 Number of words (over an alphabet of size 26) of length n with all different letters.

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A108297 Least positive k such that k * n^2 ends with n, or 0 no such k exists.

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A108343 Gapful numbers >= 100: numbers that are divisible by the number formed by their first and last digit. Numbers up to 100 trivially have this property and are excluded.

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A106710 Number of words with n letters from an alphabet of size 26 with at least two equal consecutive letters.

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