A324297 -id:A324297 - cp's OEIS frontend

A324298 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 6 (A324297).

3, 9, 15, 21, 25, 27, 33, 39, 41, 45, 51, 57, 63, 67, 69, 73, 75, 81, 87, 89, 93, 99, 105, 111, 117, 119, 121, 123, 129, 135, 137, 141, 145, 147, 153, 159, 165, 169, 171, 177, 183, 185, 189, 195, 197, 201, 207, 211, 213, 217, 219, 223, 225, 231, 233, 237, 243, 249

Offset: 1

Stefano Spezia, Mar 16 2019

All the terms of this sequence are odd.

Why? If an integer 10*k+6 = (10*a+6) * (10*b+6), then k = 10*a*b + 6*(a+b) + 3, so k is odd. - Bernard Schott, May 13 2019

			145 is a term because 26*56 = 1456 = 145*10 + 6. - _Bernard Schott_, May 13 2019

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

Cf. A017341, A053742 (ending with 5), A324297, A337856, A346389.

a={}; For[n=0,n<=250,n++,For[k=0,k<=n,k++,If[Mod[10*n+6,10*k+6]==0 && Mod[(10*n+6)/(10*k+6),10]==6 && 10*n+6>Max[10*a+6],AppendTo[a,n]]]]; a

isok6(n) = (n%10) == 6; \\ A017341
isok(k) = {my(n=10*k+6, d=divisors(n)); fordiv(n, d, if (isok6(d) && isok6(n/d), return(1))); return (0);} \\ Michel Marcus, Apr 14 2019

def aupto(lim): return sorted(set(a*b//10 for a in range(6, 10*lim//6+2, 10) for b in range(a, 10*lim//a+2, 10) if a*b//10 <= lim))
print(aupto(249)) # Michael S. Branicky, Aug 21 2021

a(n) = (A324297(n) - 6)/10.
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 1.
The conjecture is true since a(n) = (A324297(n) - 6)/10 and lim_{n->infinity} A324297(n)/A324297(n-1) = 1. - Stefano Spezia, Aug 21 2021

A346389 a(n) is the number of proper divisors of A324297(n) ending with 6.

Original entry on oeis.org

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

Offset: 1

Stefano Spezia, Jul 15 2021

			a(12) = 4 since there are 4 proper divisors of A324297(12) = 576 ending with 6: 6, 16, 36 and 96.

Cf. A017341, A032741, A324297, A324298, A337856, A346388 (ending with 5), A346392.

b={}; For[n=0, n<=450, n++, For[k=0, k<=n, k++, If[Mod[10*n+6, 10*k+6]==0 && Mod[(10*n+6)/(10*k+6), 10]==6 && 10*n+6>Max[b], AppendTo[b, 10*n+6]]]]; (* A324297 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#,10]==6&)], -1]]]]; a

a(n) = A346392(A324297(n)).

A337856 Number of positive integers with n digits that are the product of two integers ending with 6.

Original entry on oeis.org

0, 2, 20, 230, 2515, 26889, 282211, 2930013, 30196730, 309564822, 3161099901, 32182595954, 326874672928, 3313770788984

Offset: 1

Stefano Spezia, Sep 27 2020

a(n) is the number of n-digit numbers in A324297.

Index entries for sequences related to digits.

Cf. A017341, A324297, A337855, A346509.

def A337856(n):
    k, n1, n2, pset = 0, 10**(n-1)//2-18, 10**n//2-18, set()
    while 50*k**2+60*k < n2:
        a, b = divmod(n1-30*k,50*k+30)
        m = max(k,a+int(b>0))
        r = 50*k*m+30*(k+m)
        while r < n2:
            pset.add(r)
            m += 1
            r += 50*k+30
        k += 1
    return len(pset) # Chai Wah Wu, Sep 26 2021

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

a(5) corrected by and a(6)-a(9) from Jinyuan Wang, Oct 01 2020
a(10)-a(13) from Bert Dobbelaere, Oct 20 2020
a(14) from Martin Ehrenstein, Aug 06 2021

A346950 Positive integers k that are the product of two integers ending with 3.

Original entry on oeis.org

9, 39, 69, 99, 129, 159, 169, 189, 219, 249, 279, 299, 309, 339, 369, 399, 429, 459, 489, 519, 529, 549, 559, 579, 609, 639, 669, 689, 699, 729, 759, 789, 819, 849, 879, 909, 939, 949, 969, 989, 999, 1029, 1059, 1079, 1089, 1119, 1149, 1179, 1209, 1219, 1239, 1269

Offset: 1

Stefano Spezia, Aug 08 2021

All the terms end with 9 (A017377).

			9 = 3*3, 39 = 3*13, 69 = 3*23, 99 = 3*33, 129 = 3*43, 159 = 3*53, 169 = 13*13, 189 = 3*63, ...

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

Cf. A017377 (supersequence), A053742 (ending with 5), A139245 (ending with 2), A324297 (ending with 6), A346951, A346952, A346953.

a={}; For[n=0, n<=250, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+3]==0 && Mod[(10*n+9)/(10*k+3), 10]==3&& 10*n+9>Max[a], AppendTo[a, 10*n+9]]]]; a

def aupto(lim): return sorted(set(a*b for a in range(3, lim//3+1, 10) for b in range(a, lim//a+1, 10)))
print(aupto(1270)) # Michael S. Branicky, Aug 08 2021

Limit_{n->oo} a(n)/a(n-1) = 1.

A347253 Positive integers that are the product of two integers ending with 4.

Original entry on oeis.org

16, 56, 96, 136, 176, 196, 216, 256, 296, 336, 376, 416, 456, 476, 496, 536, 576, 616, 656, 696, 736, 756, 776, 816, 856, 896, 936, 976, 1016, 1036, 1056, 1096, 1136, 1156, 1176, 1216, 1256, 1296, 1316, 1336, 1376, 1416, 1456, 1496, 1536, 1576, 1596, 1616, 1656

Offset: 1

Stefano Spezia, Aug 24 2021

			16 = 4*4, 56 = 4*14, 96 = 4*24, 136 = 4*34, 176 = 4*44, 196 = 14*14, 216 = 4*54, ...

Cf. A017341 (supersequence), A053742 (ending with 5), A139245 (ending with 2), A324297 (ending with 6), A346950 (ending with 3), A347254, A347255.

a={}; For[n=0, n<=200, n++, For[k=0, k<=n, k++, If[Mod[10*n+6, 10*k+4]==0 && Mod[(10*n+6)/(10*k+4), 10]==4 && 10*n+6>Max[a], AppendTo[a, 10*n+6]]]]; a

def aupto(lim): return sorted(set(a*b for a in range(4, lim//4+1, 10) for b in range(a, lim//a+1, 10)))
print(aupto(1660)) # Michael S. Branicky, Aug 24 2021

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

A348054 Positive integers that are the product of two integers ending with 7.

Original entry on oeis.org

49, 119, 189, 259, 289, 329, 399, 459, 469, 539, 609, 629, 679, 729, 749, 799, 819, 889, 959, 969, 999, 1029, 1099, 1139, 1169, 1239, 1269, 1309, 1369, 1379, 1449, 1479, 1519, 1539, 1589, 1649, 1659, 1729, 1739, 1799, 1809, 1819, 1869, 1939, 1989, 2009, 2079, 2109

Offset: 1

Stefano Spezia, Sep 26 2021

			49 = 7*7, 119 = 7*17, 189 = 7*27, 259 = 7*37, 289 = 17*17, 329 = 7*47, 399 = 7*57, ...

Cf. A017377 (supersequence), A053742 (ending with 5), A139245 (ending with 2), A324297 (ending with 6), A346950 (ending with 3), A347253 (ending with 4), A348055.

a={}; For[n=0, n<=210, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+7]==0 && Mod[(10*n+9)/(10*k+7), 10]==7 && 10*n+9>Max[a], AppendTo[a, 10*n+9]]]]; a

def aupto(lim): return sorted(set(a*b for a in range(7, lim//7+1, 10) for b in range(a, lim//a+1, 10)))
print(aupto(2110)) # Michael S. Branicky, Sep 26 2021

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

A346507 Positive integers k that are the product of two integers greater than 1 and ending with 1.

Original entry on oeis.org

121, 231, 341, 441, 451, 561, 651, 671, 781, 861, 891, 961, 1001, 1071, 1111, 1221, 1271, 1281, 1331, 1441, 1491, 1551, 1581, 1661, 1681, 1701, 1771, 1881, 1891, 1911, 1991, 2091, 2101, 2121, 2201, 2211, 2321, 2331, 2431, 2501, 2511, 2541, 2601, 2651, 2751, 2761

Offset: 1

Stefano Spezia, Jul 21 2021

All the terms end with 1 (A017281).

			121 = 11*11, 231 = 11*21, 341 = 11*31, 441 = 21*21, 451 = 11*41, ...

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

Cf. A017281 (supersequence), A053742 (ending with 5), A324297 (ending with 6), A346508, A346509, A346510.

a={}; For[n=1, n<=300, n++, For[k=1, kMax[a], AppendTo[a, 10n+1]]]]; a

isok(k) = fordiv(k, d, if ((d>1) && (dMichel Marcus, Jul 28 2021

def aupto(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(aupto(2761)) # Michael S. Branicky, Jul 22 2021

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

The conjecture is true since it can be proved that a(n) = (sqrt(a(n-1)) + g(n-1))^2 where [g(n): n > 1] is a bounded sequence of positive real numbers. - Stefano Spezia, Aug 21 2021

A347747 Positive integers with final digit 6 that are equal to the product of two integers ending with the same digit.

Original entry on oeis.org

16, 36, 56, 96, 136, 156, 176, 196, 216, 256, 276, 296, 336, 376, 396, 416, 456, 476, 496, 516, 536, 576, 616, 636, 656, 676, 696, 736, 756, 776, 816, 856, 876, 896, 936, 976, 996, 1016, 1036, 1056, 1096, 1116, 1136, 1156, 1176, 1196, 1216, 1236, 1256, 1296, 1316

Offset: 1

Stefano Spezia, Sep 12 2021

Union of A324297 and A347253.

			16 = 4*4, 36 = 6*6, 56 = 4*14, 96 = 4*24 = 6*16, 136 = 4*34, 156 = 6*26, ...

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A017341 (supersequence), A324297, A347253, A347749.

a={}; For[n=0, n<=150, n++, For[k=0, k<=n, k++, If[Mod[10*n+6, 10*k+4]==0 && Mod[(10*n+6)/(10*k+4), 10]==4 && 10*n+6>Max[a] || Mod[10*n+6,10*k+6]==0 && Mod[(10*n+6)/(10*k+6),10]==6 && 10*n+6>Max[a], AppendTo[a, 10*n+6]]]]; a
tisdQ[n_]:=AnyTrue[{Mod[#,10],Mod[n/#,10]}&/@Divisors[n],#[[1]] == #[[2]]&]; Select[10 Range[150]+6,tisdQ] (* Harvey P. Dale, Dec 27 2021 *)

isok(m) = if ((m % 10) == 6, fordiv(m, d, if ((d % 10) == (m/d % 10), return(1)))); \\ Michel Marcus, Oct 06 2021

def aupto(lim): return sorted(set(a*b for a in range(4, lim//4+1, 10) for b in range(a, lim//a+1, 10)) | set(a*b for a in range(6, lim//6+1, 10) for b in range(a, lim//a+1, 10)))
print(aupto(1317)) # Michael S. Branicky, Sep 12 2021

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

A347748 Number of positive integers with n digits that are equal both to the product of two integers ending with 4 and to that of two integers ending with 6.

Original entry on oeis.org

0, 1, 12, 159, 1859, 20704, 223525, 2370684, 24842265, 258128126, 2665475963

Offset: 1

Stefano Spezia, Sep 12 2021

a(n) is the number of n-digit numbers in A347746.

Cf. A017341, A052268, A324297, A337856, A347253, A347255, A347746, A347749.

Table[{lo, hi}={10^(n-1), 10^n}; Length@Select[Intersection[Union@Flatten@Table[a*b, {a, 4, Floor[hi/4], 10}, {b, a, Floor[hi/a], 10}],Union@Flatten@Table[a*b, {a, 6, Floor[hi/6], 10}, {b, a, Floor[hi/a], 10}]], lo<#

def a(n):
  lo, hi = 10**(n-1), 10**n
  return len(set(a*b for a in range(4, hi//4+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi) & set(a*b for a in range(6, hi//6+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Oct 06 2021

a(n) < A052268(n).
a(n) = A337856(n) + A347255(n) - A347749(n).
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 10.

a(9)-a(10) from Michael S. Branicky, Oct 06 2021

a(11) from Frank A. Stevenson, Jan 06 2024

A347749 Number of positive integers with n digits and final digit 6 that are equal to the product of two integers ending with the same digit.

Original entry on oeis.org

0, 4, 33, 352, 3597, 36781, 374071, 3790993, 38326689, 386782889

Offset: 1

Stefano Spezia, Sep 12 2021

a(n) is the number of n-digit numbers in A347747.

Cf. A017341, A052268, A324297, A337856, A347253, A347255, A347747, A347748.

Table[{lo, hi}={10^(n-1), 10^n}; Length@Select[Union[Union@Flatten@Table[a*b, {a, 4, Floor[hi/4], 10}, {b, a, Floor[hi/a], 10}],Union@Flatten@Table[a*b, {a, 6, Floor[hi/6], 10}, {b, a, Floor[hi/a], 10}]], lo<#

def a(n):
  lo, hi = 10**(n-1), 10**n
  return len(set(a*b for a in range(4, hi//4+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi) | set(a*b for a in range(6, hi//6+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Oct 06 2021

a(n) < A052268(n).
a(n) = A337856(n) + A347255(n) - A347748(n).
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 10.

a(9)-a(10) from Michael S. Branicky, Oct 06 2021

cp's OEIS Frontend

A324298 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 6 (A324297).

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A346389 a(n) is the number of proper divisors of A324297(n) ending with 6.

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A337856 Number of positive integers with n digits that are the product of two integers ending with 6.

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A346950 Positive integers k that are the product of two integers ending with 3.

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A347253 Positive integers that are the product of two integers ending with 4.

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A348054 Positive integers that are the product of two integers ending with 7.

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A346507 Positive integers k that are the product of two integers greater than 1 and ending with 1.

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A347747 Positive integers with final digit 6 that are equal to the product of two integers ending with the same digit.