A167782 -id:A167782 - cp's OEIS frontend

A274084 Triangular numbers that are repdigits with length > 2 in some base.

15, 21, 91, 171, 666, 703, 820, 1830, 1953, 3003, 3081, 4095, 7140, 7381, 10440, 12720, 14706, 16471, 16653, 18915, 23871, 24976, 30628, 47586, 47895, 48828, 66430, 71631, 79401, 95703, 101475, 104653, 119805, 128778, 148240, 148785, 173166, 191271, 221445

Offset: 1

Robert Israel, Jun 09 2016

Intersection of A000217 and A167782.

Sequence is infinite, e.g. for any k>=2 and j>=1 it contains n*(n+1)/2 where n = ((8j+1)^k-1)/2: this has 2k digits of j in base 8j+1.

			15 = 5*6/2 = 1111_2.
21 = 6*7/2 = 111_4.
91 = 13*14/2 = 111_9.
171 = 18*19/2 = 333_7.

Robert Israel, Table of n, a(n) for n = 1..694

Cf. A000217, A045914, A167782.

N:= 10^9: # to get all entries <= N
S:= {}:
for b from 2 to floor(sqrt(N)) do
  for k from 3 do
     r:= (b^k-1)/(b-1);
     if r > N then break fi;
     for a from 1 to min(b-1, N/r) do
        if issqr(1+8*r*a) then
          S:= S union {r*a}
        fi
     od
  od
od:
sort(convert(S,list));

A290969 The least positive integer that is a repdigit with length > 2 in exactly n bases.

Original entry on oeis.org

1, 7, 31, 32767, 4095, 435356467, 16777215, 68719476735, 281474976710655

Offset: 0

Bernard Schott, Aug 16 2017

a(10) <= 1152921504606846975 = 2^60 - 1.
a(7) and following terms > 3*10^9. - Giovanni Resta, Aug 16 2017
a(7) <= 2^36-1 and a(8) <= 2^48-1. - Michel Marcus, Aug 17 2017
In fact, we have equality in both cases. - Rémy Sigrist, Aug 21 2017
Except for a(5) = (6^12 - 1) / 5, all the numbers in the data through a(8) are Mersenne numbers A000225. - Bernard Schott, Aug 27 2017

			a(1) = 7 = 111_2.
a(2) = 31 = 11111_2 = 111_5.
a(3) = 32767 = (R_15)_2 = 77777_8 = (31,31,31)_32.

Cf. A000225, A053696, A125134, A167782, A167783, A290869.

a(7) from Rémy Sigrist, Aug 19 2017

a(8) from Rémy Sigrist, Aug 21 2017

A326775 For any n >= 0, let b >= 2 be the smallest base where n has all digits equal, say to d; a(n) = d.

Original entry on oeis.org

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

Offset: 0

Rémy Sigrist, Jul 28 2019

A059711 gives base b.
From Bernard Schott, Aug 17 2019: (Start)
a(n) = 1 iff n is A220570, then n = 11_(n-1) or, n is in A053696, then n = 11..11_b for some base b.
a(n) = 2 if n = 2 * p, p prime >= 5.
a(n) = 3 if n = 3 * p, p prime >= 11.
There are k = 2 equal digits in the representation of n in the corresponding base b, except when n is a term of A167782, in which case the number k of equal digits is >= 3. (End)
n = (b^k - 1)/(b - 1) * a(n) so a(n) | n for n > 0. Furthermore a(n) <= sqrt(n). - David A. Corneth, Aug 21 2019
If b is the smallest base such that n=d*b^k+...+d*b^0 (A059711) (d=a(n) is the repdigit) then n mod b = (d*b^k+...+d*b^0) mod b = (d*b^k+...+d*b^1) mod b + (d*b^0) mod b = 0 + (d*1) mod b. Since d is less than the base we end up with the formula n mod b = d. - Jon Maiga, May 31 2021

			For n = 45:
- we have:
     b  45 in base b  Repdigit ?
     -  ------------  ----------
     2  101101        no
     3  1200          no
     4  231           no
     5  140           no
     6  113           no
     7  63            no
     8  55            yes, with d = 5
- hence a(45) = 5.

David A. Corneth, Table of n, a(n) for n = 0..9999

Cf. A059711, A053696, A060775, A220570.

a(n) = for (b=2, oo, if (#Set(digits(n,b))<=1, return (n%b)))

# with library / without (faster for large n)
from sympy.ntheory import digits
def is_repdigit(n, b): return len(set(digits(n, b)[1:])) == 1
def is_repdigit(n, b):
  if n < b: return True
  n, r = divmod(n, b)
  onlyd = r
  while n > b:
    n, r = divmod(n, b)
    if r != onlyd: return False
  return n == onlyd
def a(n):
  for b in range(2, n+3):
    if is_repdigit(n, b): return n%b
print([a(n) for n in range(87)]) # Michael S. Branicky, May 31 2021

n is a multiple of a(n).

a(n) = n mod A059711(n). - Jon Maiga, May 31 2021

A326706 Numbers m such that beta(m) = tau(m)/2 + k for some k >= 4, where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

Original entry on oeis.org

16777215, 435356467, 1073741823, 68719476735, 1099511627775, 4398046511103, 35184372088831, 281474976710655, 14901161193847656, 18014398509481983

Offset: 1

Bernard Schott, Aug 09 2019

As tau(m) = 2 * (beta(m) - k) is even, the terms of this sequence are not squares.
There are two classes of terms (see array in link and examples):
1) Non-oblong composites which have five or more Brazilian representations with three digits or more, they form a subsequence of A326705. The smallest example is a(1) = 16777215 = M_24.
2) Oblong numbers that have six or more Brazilian representations with three digits or more, they form a subsequence of A309062. The smallest example is a(9) (see 2nd example).
For a(1) to a(10), the numbers k are respectively 5, 4, 5, 6, 5, 5, 4, 7, 4 and 5.
Some Mersenne numbers are terms: M_24 = a(1), M_30 = a(3), M_36 = a(4), M_40 = a(5), M_42 = a(6), M_45 = a(7), M_48 = a(8), M_54 = a(10).

			One example of each type:
1) Non-oblong with beta"(m) = 5; tau(435356467) = 64 and 435356467 = (6^12 - 1)/5 has exactly five Brazilian representations with three digits or more: R(12)_6 = 777777_36 = (43,43,43)_216 = (259,259,259)_1296 = (31,31,31)_3747 and has 31 representations with 2 digits, so beta(435356467) = 36 and k = 4.
2) Oblong with beta"(m) = 6; tau(14901161193847656) = 768 and 14901161193847656 = (5^24 - 1)/4 = 122070312*122070313 is oblong. The six Brazilian representations with three digits or more of this term are R(24)_5 = 666666666666_25 = (31,31,31,31,31,31,31,31)_125 = (156,156,156,156,156)_625, =(3906,3906,3906,3906)_15625 = (97656,97656,97656)_390625 so beta"(14901161193847656) = 6 and beta(61035156) = (tau(61035156)/2 - 2) + 6 = 388 and k = 4.

Bernard Schott, Array of relations beta = f(tau)
Index entries for sequences related to Brazilian numbers

Cf. A000005 (tau), A220136 (beta).
Subsequence of A167782, A167783 and A290869.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3), this sequence (tau(m)/2 + k, k >= 4).
Cf. A291592 (Mersenne numbers).

okrepu3(b, target, lim) = {my(k = 3, nb = 0, x); while ((x=(b^k-1)/(b-1)) <= target, if (x==target, nb++); k++); nb; }
dge3(n, d) = {my(nb=0, ndi, limi); for (i=1, #d, ndi = n/d[i]; limi = sqrtint(ndi); for (k=d[i]+1, limi, nb += okrepu3(k, ndi, limi); ); ); nb; }
deq2(n, d) = {my(nb=0, nk); for (k=1, #d\2, nk = (n - d[k])/d[k]; if (nk > d[k], nb++); ); nb; }
beta(n) = {if (n<3, return (0)); my(d=divisors(n)); deq2(n, d) + dge3(n, d) - 1; }
isok(n) = beta(n) - numdiv(n)/2 > = 4; \\ Michel Marcus, Aug 10 2019

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A274084 Triangular numbers that are repdigits with length > 2 in some base.

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A290969 The least positive integer that is a repdigit with length > 2 in exactly n bases.

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A326775 For any n >= 0, let b >= 2 be the smallest base where n has all digits equal, say to d; a(n) = d.

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A326706 Numbers m such that beta(m) = tau(m)/2 + k for some k >= 4, where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

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