A053742 -id:A053742 - cp's OEIS frontend

A346388 a(n) is the number of proper divisors of A053742(n) ending with 5.

1, 3, 2, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 7, 3, 3, 7, 5, 3, 7, 3, 3, 8, 3, 5, 7, 3, 5, 7, 3, 3, 11, 5, 3, 7, 3, 3, 7, 7, 3, 9, 3, 5, 7, 3, 7, 7, 5, 3, 11, 3, 3, 11, 3, 3, 7, 3, 5, 11, 7, 5, 7, 4, 3, 7, 3, 7, 11, 3, 3, 7, 7, 5, 11, 3, 3, 11, 5, 3, 7, 7, 3, 11, 3, 5

Offset: 0

Stefano Spezia, Jul 15 2021

			a(10) = 7 since there are 7 proper divisors of A053742(10) = 525 ending with 5: 5, 15, 25, 35, 75, 105 and 175.

Cf. A032741, A053742, A346389 (ending with 6), A346392.

a[n_]:=Length[Drop[Select[Divisors[25+50n], (Mod[#,10]==5&)], -1]]; Array[a, 90, 0]

a(n) = A346392(A053742(n)).

A016873 a(n) = 5*n + 2.

Original entry on oeis.org

2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, 102, 107, 112, 117, 122, 127, 132, 137, 142, 147, 152, 157, 162, 167, 172, 177, 182, 187, 192, 197, 202, 207, 212, 217, 222, 227, 232, 237, 242, 247, 252, 257, 262, 267, 272, 277

Offset: 0

N. J. A. Sloane

Numbers ending in 2 or 7. - Lekraj Beedassy, Jul 08 2006
For n > 2, also the number of (not necessarily maximal) cliques in the n-gear graph. - Eric W. Weisstein, Nov 29 2017
Also, positive integers k such that 10*k+5 is equal to the product of two integers ending with 5. Proof: if 10*k+5 = (10*a+5) * (10*b+5), then k = 10*a*b + 5*(a+b) + 2 = 5 * (a + b + 2*a*b) + 2, of the form 5m + 2. So, 262 is a term because 2625 = 35 * 75. - Bernard Schott, May 15 2019
Numbers k such that 2^x + 3^x == 0 mod 31 and 2^x + 3^x == 0 mod 11 with x = 6*k+3. - Pedro Caceres, May 18 2022

G. C. Greubel, Table of n, a(n) for n = 0..1000
Cino Hilliard, solutions to 3^x + 5^x == 2 mod 11. [broken link]
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A001622, A008586, A008587, A016861, A342757.
Cf. A053742 (product of two integers ending with 5).
Cf. A324298 (product of two integers ending with 6).

[5*n+2: n in [0..80]]; // G. C. Greubel, Oct 17 2023

a[1]:=2:for n from 2 to 100 do a[n]:=a[n-1]+5 od: seq(a[n], n=1..50); # Zerinvary Lajos, Mar 16 2008

Range[2, 500, 5] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
(* Programs from Eric W. Weisstein, Nov 29 2017 *)
5*Range[0, 70] +2
LinearRecurrence[{2, -1}, {7, 12}, {0, 70}]
CoefficientList[Series[(2+3*x)/(1-x)^2, {x,0,70}], x] (* End *)

a(n)=5*n+2 \\ Charles R Greathouse IV, Jul 10 2016

[i+2 for i in range(300) if gcd(i,5) == 5] # Zerinvary Lajos, May 20 2009

a(n) = 10*n - a(n-1) - 1 (with a(0)=2). - Vincenzo Librandi, Nov 20 2010
G.f.: (2+3*x)/(1-x)^2. - Colin Barker, Jan 08 2012
E.g.f.: exp(x)*(2 + 5*x). - Stefano Spezia, Mar 21 2021
Sum_{n>=0} (-1)^n/a(n) = sqrt(2-2/sqrt(5))*Pi/10 + log(phi)/sqrt(5) - log(2)/5, where phi is the golden ratio (A001622). - Amiram Eldar, Apr 15 2023

A324297 Positive integers k that are the product of two integers ending with 6.

Original entry on oeis.org

36, 96, 156, 216, 256, 276, 336, 396, 416, 456, 516, 576, 636, 676, 696, 736, 756, 816, 876, 896, 936, 996, 1056, 1116, 1176, 1196, 1216, 1236, 1296, 1356, 1376, 1416, 1456, 1476, 1536, 1596, 1656, 1696, 1716, 1776, 1836, 1856, 1896, 1956, 1976, 2016, 2076, 2116

Offset: 1

Stefano Spezia, Mar 16 2019

All the terms end with 6 (A017341).

			36 = 6*6, 96 = 6*16, 216 = 6*36, 256 = 16*16, 276 = 6*46, ...

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

Cf. A000400, A017341 (supersequence), A324298, A053742 (ending with 5).

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[a],AppendTo[a,10*n+6]]]]; a

isok6(n) = (n%10) == 6; \\ A017341
isok(n) = {if (isok6(n), my(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 for a in range(6, lim//6+1, 10) for b in range(a, lim//a+1, 10)))
print(aupto(2117)) # Michael S. Branicky, Aug 18 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 18 2021

A016850 a(n) = (5*n)^2.

Original entry on oeis.org

0, 25, 100, 225, 400, 625, 900, 1225, 1600, 2025, 2500, 3025, 3600, 4225, 4900, 5625, 6400, 7225, 8100, 9025, 10000, 11025, 12100, 13225, 14400, 15625, 16900, 18225, 19600, 21025, 22500, 24025, 25600, 27225, 28900, 30625, 32400, 34225, 36100, 38025, 40000, 42025

Offset: 0

N. J. A. Sloane

If we define C(n) = (5*n)^2 (n > 0), the sequence is the first "square-sequence" such that for every n there exists p such that C(n) = C(p) + C(p+n). We observe in fact that p = 3*n because 25 = 3^2 + 4^2. The sequence without 0 is linked with the first nontrivial solution (trivial: n^2 = 0^2 + n^2) of the equation X^2 = 2*Y^2 + 2*n^2 where X = 2*k and Y = 2*p + n which is equivalent to k^2 = p^2 + (p+n)^2 for n given. The second such "square-sequence" is (29*n)^2 (n > 0) because 29^2 = 20^2 + 21^2 and with this relation we obtain (29*n)^2 = (20*n)^2 + (20*n+n)^2. - Richard Choulet, Dec 23 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A033429, A053742 (first differences), A008587, A008607.

Similar sequences listed in A244630.

[(5*n)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

(5Range[0, 31])^2 (* Alonso del Arte, Oct 08 2017 *)

a(n)=(5*n)^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 25*n^2 = 25*A000290(n) = 5*A033429(n). - Omar E. Pol, Jul 03 2014
From Amiram Eldar, Jan 25 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/150.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/300.
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/5)/(Pi/5).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/5)/(Pi/5) = 5*sqrt((5-sqrt(5))/2)/(2*Pi). (End)
a(n) = Sum_{i=0..n-1} A053742(i). - John Elias, Jun 30 2021
G.f.: 25*x*(1 + x)/(1 - x)^3. - Stefano Spezia, Jul 08 2023
From Elmo R. Oliveira, Nov 30 2024: (Start)
E.g.f.: 25*x*(1 + x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = n*A008607(n) = A000290(A008587(n)) = A008587(n)^2. (End)

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.

A132359 Numbers divisible by the square of their last decimal digit.

Original entry on oeis.org

1, 11, 12, 21, 25, 31, 32, 36, 41, 51, 52, 61, 63, 64, 71, 72, 75, 81, 91, 92, 101, 111, 112, 121, 125, 128, 131, 132, 141, 144, 147, 151, 152, 153, 161, 171, 172, 175, 181, 191, 192, 201, 211, 212, 216, 221, 224, 225, 231, 232, 241, 243, 251, 252, 261, 271, 272

Offset: 1

Jonathan Vos Post, Nov 08 2007

Subsequences are A017281 and A053742 representing last digits 1 and 5. Generators for the subsequences representing last digits 2, 3, 4, 6, 7, 8 and 9 are, in that order, the terms 12+20i, 63+90i, 64+80i, 36+180i, 147+490i, 128+320i, 729+810i, where i=0,1,2,... - R. J. Mathar, Nov 13 2007

This is a 10-automatic sequence. - Charles R Greathouse IV, Dec 28 2011

			147 belongs to the sequence because 147/7^2 = 3.

T. D. Noe and Christian N. K. Anderson, Table of n, a(n) for n = 1..10000 (first 1000 values from T. D. Noe)
Index entries for 10-automatic sequences.

Cf. A034709, A225722, A221651, A225296, A225297, A034709, A034837, A005349, A007602, A034838, A225297.

isA132359 := proc(n) local ldig ; ldig := n mod 10 ; if ldig <> 0 and n mod (ldig^2) = 0 then true ; else false ; fi ; end: for n from 1 to 400 do if isA132359(n) then printf("%d,",n) ; fi ; od: # R. J. Mathar, Nov 13 2007
a:=proc(n) local nn: nn:=convert(n,base,10): if 0 < nn[1] and `mod`(n,nn[1]^2) =0 then n else end if end proc: seq(a(n),n=1..250); # Emeric Deutsch, Nov 15 2007

Select[Range[250], IntegerDigits[ # ][[ -1]] > 0 && Mod[ #, IntegerDigits[ # ][[ -1]]^2] == 0 &] (* Stefan Steinerberger, Nov 12 2007 *)
dsldQ[n_]:=Module[{lidnsq=Last[IntegerDigits[n]]^2},lidnsq!=0 && Divisible[n,lidnsq]]; Select[Range[300],dsldQ] (* Harvey P. Dale, May 03 2011 *)

is(n)=n%(n%10)^2==0 \\ Charles R Greathouse IV, Dec 28 2011

def ok(n): return n%10 > 0 and n%(n%10)**2 == 0
print([k for k in range(273) if ok(k)]) # Michael S. Branicky, Jul 03 2022

which(sapply(1:500,function(x) isint(x/(x%%10)^2))) # Christian N. K. Anderson, May 04 2013

Numbers k such that fp[k / (k mod 10)] = 0.

a(n) ~ 6350400*n/1241929 = 5.113...*n. - Charles R Greathouse IV, Dec 28 2011

Corrected and extended by Stefan Steinerberger, Emeric Deutsch and R. J. Mathar, Nov 12 2007

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

Original entry on oeis.org

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

A337855 Number of n-digit positive integers that are the product of two integers ending with 5.

Original entry on oeis.org

0, 2, 18, 180, 1800, 18000, 180000, 1800000, 18000000, 180000000, 1800000000, 18000000000, 180000000000, 1800000000000, 18000000000000, 180000000000000, 1800000000000000, 18000000000000000, 180000000000000000, 1800000000000000000, 18000000000000000000, 180000000000000000000

Offset: 1

Stefano Spezia, Sep 27 2020

a(n) is the number of n-digit numbers in A053742.

Christian Krause, LODA, an assembly language, a computational model and a tool for mining integer sequences
Index entries for linear recurrences with constant coefficients, signature (10).
Index entries for sequences related to digits.

Cf. A011557 (powers of 10), A052268 (number of n-digit integers), A053742 (product of two integers ending with 5), A093136, A337856.

Mathematica
```
LinearRecurrence[{10},{0,0,2,18},22]
```

O.g.f.: 2*(1 - x)*x^2/(1 - 10*x).
E.g.f.: (9*exp(10*x) - 9 - 90*x + 50*x^2)/500.
a(n) = 10*a(n-1) for n > 3 , with a(1) = 0, a(2) = 2 and a(3) = 18.
a(n) = 18*10^(n-3) for n > 2.
a(n) = 18*A011557(n - 3) for n > 2.
a(n) = 2*A052268(n - 2) for n > 2.
Sum_{i=2..n} a(n) = A093136(n - 1) for n > 1.
a(n) = 2*floor((k + 27*10^(n-2))/30), with 2 < k < 28. [This formula was found in the form k = 7 by Christian Krause's LODA miner] - Stefano Spezia, Dec 06 2021

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.

cp's OEIS Frontend

A346388 a(n) is the number of proper divisors of A053742(n) ending with 5.

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A016873 a(n) = 5*n + 2.

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A324297 Positive integers k that are the product of two integers ending with 6.

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A016850 a(n) = (5*n)^2.

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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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A132359 Numbers divisible by the square of their last decimal digit.

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