A346507 -id:A346507 - cp's OEIS frontend

A346508 Positive integers k such that 10*k+1 is equal to the product of two integers greater than 1 and ending with 1 (A346507).

12, 23, 34, 44, 45, 56, 65, 67, 78, 86, 89, 96, 100, 107, 111, 122, 127, 128, 133, 144, 149, 155, 158, 166, 168, 170, 177, 188, 189, 191, 199, 209, 210, 212, 220, 221, 232, 233, 243, 250, 251, 254, 260, 265, 275, 276, 282, 287, 291, 296, 298, 309, 311, 313, 317

Offset: 1

Stefano Spezia, Jul 21 2021

			107 is a term because 21*51 = 1071 = 107*10 + 1.

Stefano Spezia, Table of n, a(n) for n = 1..10000

Cf. A016873 (ending with 5), A017281, A324298 (ending with 6), A346507, A346509, A346510.

a={}; For[n=1, n<=350, n++, For[k=1, kMax[10a+1], AppendTo[a, n]]]]; a

def aupto(lim): return sorted(set(a*b//10 for a in range(11, 10*lim//11+2, 10) for b in range(a, 10*lim//a+2, 10) if a*b//10 <= lim))
print(aupto(318)) # Michael S. Branicky, Aug 21 2021

a(n) = (A346507(n) - 1)/10.
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 1.
The conjecture is true since a(n) = (A346507(n) - 1)/10 and lim_{n->infinity} A346507(n)/A346507(n-1) = 1. - Stefano Spezia, Aug 21 2021

A346510 a(n) is the number of nontrivial divisors of A346507(n) ending with 1.

Original entry on oeis.org

1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 4, 2, 4, 2

Offset: 1

Stefano Spezia, Jul 21 2021

			a(42) = 4 since there are 4 nontrivial divisors of A346507(42) = 2541 ending with 1: 11, 21, 121 and 231.

Cf. A017281, A070824, A346388 (ending with 5), A346389 (ending with 6), A346392, A346507, A346508, A346509.

b={}; For[n=1, n<=500, n++, For[k=1, kMax[b], AppendTo[b, 10n+1]]]]; (* A346507 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#, 10]==1&)], -1]]-1]]; a

f(n) = sumdiv(n, d, (d>1) && (d(f(x)), [1..5000])) \\ Michel Marcus, Jul 28 2021

from sympy import divisors
def f(n): return sum(d%10 == 1 for d in divisors(n)[1:-1])
def A346507upto(lim): return sorted(set(a*b for a in range(11, lim//11+1, 10) for b in range(a, lim//a+1, 10)))
print(list(map(f, A346507upto(5000)))) # Michael S. Branicky, Jul 31 2021

a(n) = A346392(A346507(n)) - 1.

A346509 Number of positive integers with n digits that are the product of two integers greater than 1 and ending with 1.

Original entry on oeis.org

0, 0, 12, 200, 2660, 31850, 361985, 3982799, 42914655, 455727689, 4788989458, 49930700093, 517443017072, 5336861879564

Offset: 1

Stefano Spezia, Jul 21 2021

a(n) is the number of n-digit numbers in A346507.

Cf. A017281, A052268, A087630, A337855 (ending with 5), A337856 (ending with 6), A346507.

a(n) = {my(res = 0); forstep(i = 10^(n-1) + 1, 10^n, 10, f = factor(i); if(bigomega(f) == 1, next); d = divisors(f); for(j = 2, (#d~ + 1)>>1, if(d[j]%10 == 1 && d[#d + 1 - j]%10 == 1, res++; next(2) ) ) ); res } \\ David A. Corneth, Jul 22 2021

def A346507upto(lim): return sorted(set(a*b for a in range(11, lim//11+1, 10) for b in range(a, lim//a+1, 10)))
def a(n): return len(A346507upto(10**n)) - len(A346507upto(10**(n-1)))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Jul 22 2021

Conjecture: Lim_{n->infinity} a(n)/a(n-1) = 10.

a(6)-a(9) from Michael S. Branicky, Jul 22 2021
a(10) from David A. Corneth, Jul 22 2021
a(11) from Michael S. Branicky, Jul 23 2021
a(11) corrected and extended with a(12) by Martin Ehrenstein, Aug 03 2021
a(13)-a(14) from Martin Ehrenstein, Aug 05 2021

A346526 Positive integers k that are the product of two integers greater than 1 and ending with the same digit as k.

Original entry on oeis.org

25, 36, 75, 96, 100, 121, 125, 156, 175, 200, 216, 225, 231, 256, 275, 276, 300, 325, 336, 341, 375, 396, 400, 416, 425, 441, 451, 456, 475, 500, 516, 525, 561, 575, 576, 600, 625, 636, 651, 671, 675, 676, 696, 700, 725, 736, 756, 775, 781, 800, 816, 825, 861, 875

Offset: 1

Stefano Spezia, Jul 22 2021

Union of 100*A000027, A053742, A324297 and A346507.

			25 = 5*5, 36 = 6*6, 75 = 5*15, 96 = 6*16, 100 = 10*10, 121 = 11*11, 125 = 5*25, 156 = 6*26, 175 = 5*35, 200 = 10*20, 216 = 6*36, 225 = 15*15, 231 = 11*21, ...

Fung Cheok Yin, Limit of the ratio of a(n) to a(n-1) in A346526, Aug 12 2021.

Cf. A000027, A053742, A324297, A346507.

(setf candidates (list 25)) (setf result nil)
(defun factor (num small-num) (equalp 0 (mod num small-num)))
(defun same-end-digit (num1 num2 num3) (and (equalp (mod num1 10) (mod num2 10)) (equalp (mod num2 10) (mod num3 10))))
(defun good-factor-p (num) (loop for i from 5 to (sqrt num) do ( if (factor num i) ( if (same-end-digit num i (/ num i) ) (return T) ))))
(loop for i from 26 to 9000 do ( if (or (equalp 0 (mod i 10)) (equalp 1 (mod i 10)) (equalp 5 (mod i 10)) (equalp 6 (mod i 10))) (push i candidates)))
(dolist (element candidates) (if (good-factor-p element) (push element result)))
(format t (write-to-string result)) \\ FUNG Cheok Yin, Aug 12 2021

isok(k) = my(u=k%10); sumdiv(k, d, (d>1) && (d 0; \\ Michel Marcus, Jul 23 2021

Lim_{n->infinity} a(n)/a(n-1) = 1.

cp's OEIS Frontend

A346508 Positive integers k such that 10*k+1 is equal to the product of two integers greater than 1 and ending with 1 (A346507).

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A346510 a(n) is the number of nontrivial divisors of A346507(n) ending with 1.

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A346509 Number of positive integers with n digits that are the product of two integers greater than 1 and ending with 1.

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A346526 Positive integers k that are the product of two integers greater than 1 and ending with the same digit as k.

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