Reiner Moewald - cp's OEIS frontend

A367378 2*k-digit squares with the left half being a reversed k-digit square and the right half being a k-digit square.

49, 1849, 144400, 148225, 522729, 16564900, 40322500, 46717225, 98446084, 98863249, 1429293636, 1697440000, 4013222500, 4228250625, 4247128900, 4684991809, 5205622500, 5227290000, 6161465025, 6557274529, 104409765625, 121975562500, 123151864900, 127413302500, 140301186624

Offset: 1

Reiner Moewald, Nov 15 2023

			522729 is in the sequence since reversed the left half is the square 15^2 and the right half is the square 27^2.
A 6-digit term might start with 522 as 522 is the reversal of a three-digit square (namely 225 = 15^2). If a 6-digit term starts with 522 then it is between (inclusive) 522100 and 522999. The only such square is 522729. As 729 (= 522729 - 522000) is a square we have 522729 is in the sequence. - _David A. Corneth_, Nov 21 2023

Michael S. Branicky, Table of n, a(n) for n = 1..1031 (all terms <= 40 digits)
David A. Corneth, PARI program

Cf. A000290, A004086.

PARI
```
\\ See PARI link
```

from math import isqrt
from itertools import count, islice
def agen(): # generator of terms
    for k in count(1):
        lb, ub, sk = isqrt(10**(k-1)-1)+1, isqrt(10**k-1), set()
        for i in range(lb, ub+1):
            if i%10 == 0: continue
            left = int(str(i*i)[::-1]) * 10**k
            # loop below based on idea by David A. Corneth in Example
            lbt, ubt = isqrt(left-1)+1, isqrt(left + 10**k - 1)
            for t in range(lbt, ubt+1):
                tt = t*t
                right = tt - left
                sr = str(right)
                if len(sr) == k and isqrt(right)**2 == right:
                    sk.add(tt)
        yield from sorted(sk)
print(list(islice(agen(), 20))) # Michael S. Branicky, Nov 21 2023

More terms from David A. Corneth, Nov 20 2023

A356334 a(n) is the number of nonnegative integer solutions (x; y) with x <= y of x^(n+1) + y^(n+1) = (x+y)^n.

Original entry on oeis.org

1, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 0

Reiner Moewald, Aug 04 2022

Conjecture: a(n) = 3 except for n = 0 or n = 2. - Chai Wah Wu, Sep 21 2022

			For n=2, the 4 solutions are (0; 0), (0; 1), (1; 2) and (2; 2). The pairs (0; 0), (0; 1) and (2^(n-1); 2^(n-1)) exist for all n > 0.

def A356334(n): return sum(1 for x in range((1<Chai Wah Wu, Sep 19 2022

a(11)-a(14) from Hugo Pfoertner, Sep 18 2022

a(15)-a(20) from Chai Wah Wu, Sep 21 2022

A347189 Positive integers that can be expressed as the sum of powers of their digits (from right to left) with consecutive natural exponents.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 24, 332, 1676, 121374, 4975929, 134116265, 1086588775, 3492159897, 8652650287, 8652650482

Offset: 1

Reiner Moewald, Aug 21 2021

Any number > 9 consisting of just one digit (A014181) can't be in the list. (Provable using the Carmichael function.)

			24 = 4^2 + 2^3 is a term.
332 = 2^3 + 3^4 + 3^5 is another term.

Cf. A023052, A032799, A014181.

liste = []
for ex in range(0, 20):
    for t in range(1, 10000):
        n = t
        pot = ex
        ergebnis = 0
        while n > 0:
            pot = pot + 1
            rest = n % 10
            n = (n - rest) // 10
            zw = 1
            for i in range(pot):
                zw = zw * rest
            ergebnis = ergebnis + zw
        if (int(ergebnis) == t) and (t not in liste):
            liste.append(t)
liste.sort()
print(liste)

def powsum(digits, startexp):
    return sum(digits[i]**(startexp+i) for i in range(len(digits)))
def ok(n):
    if n < 10: return True
    s = str(n)
    if set(s) <= {'0', '1'}: return False
    digits, startexp = list(map(int, s))[::-1], 1
    while powsum(digits, startexp) < n: startexp += 1
    return n == powsum(digits, startexp)
print(list(filter(ok, range(2*10**5)))) # Michael S. Branicky, Aug 29 2021

a(15)-a(19) from Michael S. Branicky, Aug 31 2021

A344633 Lengths k of k-digit integers of the form 1, 12, 123, 1234, ... (A057137) which are divisible by k.

Original entry on oeis.org

1, 2, 3, 5, 6, 9, 10, 12, 14, 15, 16, 18, 30, 90, 96, 110, 197, 210, 270, 330, 390, 410, 630, 810, 930, 959, 990, 1110, 1170, 1210, 1230, 1470, 1710, 1890, 1956, 2310, 2430, 2530, 2538, 2710, 2730, 2790, 2802, 2922, 2970, 3330, 3510, 3519, 3630, 3690, 4115, 4245

Offset: 1

Reiner Moewald, May 26 2021

It is easy to prove that 10*3^k, k >= 0 is always a solution.

			3 is a term since 123 is divisible by 3 (123 = 3*41).

Cf. A057137.

for n from 1 to 5000 do
    if modp(A057137(n),n) = 0 then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Aug 16 2021

f(n) = 137174210*10^n\1111111111; \\ A057137
isok(k) = (f(k) % k) == 0; \\ Michel Marcus, Aug 16 2021

a ="1234567890"
for k in range(10):
    a = a + a
sol = ""
for n in range(1, len(a)):
    if int(a[0:n]) % n == 0:
        sol = sol + str(n) + ", "
print(sol)

{n: n|A057137(n)}. - R. J. Mathar, Aug 16 2021

A334929 Positive integers k such that there exists a positive integer m consisting of k identical digits and such that m is a multiple of k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, 22, 24, 27, 36, 42, 44, 45, 54, 63, 66, 72, 78, 81, 84, 88, 108, 111, 126, 132, 135, 156, 162, 168, 189, 198, 205, 216, 222, 234, 242, 243, 252, 264, 294, 312, 324, 333, 342, 378, 396, 404, 405, 444, 462, 465, 468, 484, 486

Offset: 1

Reiner Moewald, May 16 2020

For k=3^t, t>=1 you can always find numbers m.

			12 is a term since 444444444444 = 12*37037037037.

Cf. A014950.

ok[n_] := AnyTrue[(10^n - 1)/9 Range@9, Mod[#, n] == 0 &]; Select[ Range[486], ok] (* Giovanni Resta, May 24 2020 *)

t = "1"
list = [1]
for i in range(1, 1000):
    t = "1" + t
    m = int(t)
    weiter = 0
    for k in range(1, 10):
        if k * m % (i + 1) == 0:
            weiter = 1
    if weiter == 1:
        list.append(i + 1)
print(list)

A333312 Positive integers m >= 2 such that there is a value of v such that the sequence {v, ..., v + m - 1} of m nonnegative integers can be partitioned into two subsets of consecutive integers with the same sum.

Original entry on oeis.org

9, 15, 20, 21, 24, 25, 27, 28, 33, 35, 36, 39, 40, 44, 45, 48, 49, 51, 52, 55, 56, 57, 60, 63, 65, 68, 69, 72, 75, 76, 77, 80, 81, 84, 85, 87, 88, 91, 92, 93, 95, 96, 99, 100, 104, 105, 108, 111, 112, 115, 116, 117, 119, 120, 121, 123, 124, 125, 129, 132, 133

Offset: 1

Reiner Moewald, Mar 14 2020

nonn

There are no subsets for m = 4 * k + 6 and just one subset for prime numbers m.

			m = 9, v=16: 16 + 17 + 18 + 19 + 20 = 21 + 22 + 23 + 24.
m = 9, v=2: 2 + 3 + 4 + 5 + 6 + 7 = 8 + 9 + 10 = 27.

Wilfried Haag, Die Wurzel. Problem 2020 - 14. (March/April 2020: www.wurzel.org)

for m in range(3, 1000):
    anz = 0
    for i in range(m // 2 + 1, m):
        l = (2 * i * i - 2 * i - m * (m - 1)) / (2 * (m - 2 * i))
        if l - int(l) == 0 and l >= 0:
            anz = anz + 1
    if anz > 1:
        print(m)

For m = 6 * k + 3, you can always find two subsets of {v, ...,v+m-1} of length 3 * k + 2 with v = (3 * k + 1)^2 and length 3 * k + 3 with v = 3*k^2-1 elements.

A328343 Numbers k such that it is possible to find k consecutive squares whose sum is equal to the sum of two consecutive squares.

Original entry on oeis.org

1, 2, 3, 10, 17, 25, 26, 34, 41, 50, 51, 65, 73, 82, 89, 97, 106, 113, 122, 123, 145, 146, 169, 170, 178, 185, 194, 218, 219, 241, 250, 257, 267, 274, 281, 291, 298, 305, 314, 338, 339, 353, 362, 370, 377, 386, 394, 401, 409, 410, 411, 433, 449, 457, 505, 530, 545

Offset: 1

Reiner Moewald, Oct 13 2019

nonn

The program generates solutions to the problem, but not necessarily all solutions. To exclude the existence of further solution you have to apply coset arithmetics (modulo operations).

			k = 1: 3^2 + 4^2 = 5^2.
k = 2: 3^2 + 4^2 = 3^2 + 4^2.
k = 3: 13^2 + 14^2 = 10^2 + 11^2 + 12^2.
k = 10: 26^2 + 27^2 = 7^2 + ... + 16^2.
k = 17: 40^2 + 41^2 = 5^2 + ... + 21^2.
k = 25: 78^2 + 79^2 = 9^2 + ... + 33^2.
k = 26: 205^2 + 206^2 = 44^2 + ... + 49^2.
k = 34: 856^2 + 857^2 = 191^2 + ... + 224^2.
k = 41: 3029^2 + 3030^2 = 649^2 + ... + 689^2.
k = 50: 146^2 + 147^2 = 1^2 + ... + 50^2.
k = 51: 210^2 + 211^2 = 14^2 + ... + 64^2.
k = 65: 236^2 + 237^2 = 5^2 + ... + 69^2.
k = 73: 278^2 + 279^2 = 5^2 + ... + 76^2.
k = 82: 1070^2 + 1071^2 = 125^2 + ... + 206^2.
k = 89: 147445^2 + 147446^2 = 22059^2 + ... + 22147^2.
k = 97: 544^2 + 545^2 = 25^2 + ... + 121^2.

Cf. A001032, A185545.

Select[Range[60], {} != FindInstance[ Sum[t^2, {t, x, x+#-1}] == y^2 + (y + 1)^2, {x, y}, Integers] &] (* Giovanni Resta, Oct 23 2019 *)

import math
for n in range(1, 100):
   for b in range(1, 10000000):
      d = (6*b*b*(n+1)+6*b*n*(n+1)+2*n*n*n+3*n*n+n)
      w = int((math.sqrt(d/6)))
      a = w
      if 6*a*a-6*b*b*(n+1)-6*b*n*(n+1)-2*n*n*n-3*n*n-n == 0:
         print(a,b,n+1)
      a = w+1
      if 6*a*a-6*b*b*(n+1)-6*b*n*(n+1)-2*n*n*n-3*n*n-n == 0:
         print(a,b,n+1)
      a = w-1
      if 6*a*a-6*b*b*(n+1)-6*b*n*(n+1)-2*n*n*n-3*n*n-n == 0:
         print(a,b,n+1)

a(17)-a(38) from Jon E. Schoenfield, Oct 22 2019

a(39)-a(57) from Jon E. Schoenfield and Giovanni Resta, Oct 23 2019

A309417 Number of steps needed to reduce 10^n to zero by subtracting its digital sum.

Original entry on oeis.org

2, 11, 81, 611, 4798, 39320, 333583, 2897573, 25632474, 230231687, 2091437006, 19145032382, 176258021378, 1630867803755, 15161044498785, 141573907590908, 1327916557473475, 12513166293358138, 118472791400037286, 1126683083504083356, 10754171449735292485

Offset: 1

Reiner Moewald, Jul 30 2019

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

			a(100)=11 since 100->99->81->72->63->54->45->36->27->18->9->0.

Dominic McCarty, Table of n, a(n) for n = 1..500
Dominic McCarty, Python program for A309417

Cf. A066568 (n - sum of digits of n).

f[n_] := Length[NestWhileList[# - Total[IntegerDigits[#]]&, n, # > 0 &]]-1; f /@ (10^Range[8]) (* Amiram Eldar, Aug 08 2019 *)

a(n)={my(s=10^n, k=0); while(s, k++; s-=sumdigits(s)); k} \\ Andrew Howroyd, Sep 09 2019

import math
def digitsum(n):
   ds = 0
   while n > 0:
      ds += n % 10
      n = n // 10
   return ds
def steps(n):
   count = 0
   while n > 0:
      n = n - digitsum(n)
      count += 1
   return count
n = 1
for i in range(1,10):
   n = 10 * n
   print(steps(n))

a(13)-a(15) from Giovanni Resta, Sep 10 2019

a(16) and on from Dominic McCarty, Feb 12 2025

A323612 Numbers k such that k^2 starts with at least the same number >= 2 of equal digits as does k itself.

Original entry on oeis.org

88, 332, 334, 335, 336, 337, 338, 339, 664, 665, 667, 668, 669, 880, 881, 882, 883, 995, 996, 997, 998, 3317, 3318, 3319, 3320, 3321, 3322, 3323, 3324, 3325, 3326, 3327, 3328, 3329, 3332, 3334, 3335, 3336, 3337, 3338, 3339, 3340, 3341, 3342, 3343, 3344, 3345, 3346, 3347

Offset: 1

Reiner Moewald, Jan 20 2019

			332^2 = 110224, i.e., both the integer and its square start with two equal digits.
Also 334^2 = 111556, so 334 is a term.

Reiner Moewald, Table of n, a(n) for n = 1..24527

nbi(d) = my(nb=1); for(k=2, #d, if (d[k] == d[1], nb++, break)); nb;
isok(n) = my(nb = nbi(digits(n))); (nb > 1) && (nbi(digits(n^2)) >= nb); \\ Michel Marcus, Apr 24 2019

def zahl(z):
   a = str(z)
   r = 0
   while r < len(a) and a[0] == a[r]:
      r = r + 1
   b = str(z*z)
   s = 0
   while r < len(b) and b[0] == b[s]:
      s = s + 1
   if r >= 2:
      return(s-r)
   else:
      return(-1)
anz = 0
for i in range(1000000):
   if zahl(i) >= 0:
      anz = anz +1
      print(anz, i)

A308391 Number of ordered pairs of n-digit positive integers the product of which is a 2n-digit integer.

Original entry on oeis.org

58, 6610, 668843, 66965113, 6697324753, 669740590290, 66974140069358, 6697414817000983, 669741489800555031, 66974149061059480123

Offset: 1

Reiner Moewald, May 23 2019

			a(1)=58 since we get the following pairs: (2, 5), ..., (2, 9), (3, 4), ..., (3, 9), (4, 3), ..., (4, 9), (5, 2), ..., (5, 9), (6, 2), ..., (6, 9), (7, 2), ..., (7, 9), (8, 2), ..., (8, 9), (9, 2), ..., (9, 9).

Cf. A174425.

a(n) = 9*10^(2*n-1) - 10^n - sum(k=10^(n-1)+1, 10^n-1, ceil(10^(2*n-1)/k)); \\ Michel Marcus, Jun 25 2019

import math
ende = 1
for i in range(1,10):
   anz = 0
   for a in range(ende, 10*ende):
      z = math.ceil((ende*ende*10)/a)
      if z < ende*10:
         anz = anz + ende*10 - z
   ende = ende*10
   print(i, anz)

a(n) = 9*10^(2n-1) - 10^n - Sum_{k=10^(n-1)+1..10^n-1} ceiling(10^(2n-1)/k).

a(n) ~ (9-log(10))*10^(2n-1).

cp's OEIS Frontend

User: Reiner Moewald

A367378 2*k-digit squares with the left half being a reversed k-digit square and the right half being a k-digit square.

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PARI

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A356334 a(n) is the number of nonnegative integer solutions (x; y) with x <= y of x^(n+1) + y^(n+1) = (x+y)^n.

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A347189 Positive integers that can be expressed as the sum of powers of their digits (from right to left) with consecutive natural exponents.

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Python

Python

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A344633 Lengths k of k-digit integers of the form 1, 12, 123, 1234, ... (A057137) which are divisible by k.

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Maple

PARI

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A334929 Positive integers k such that there exists a positive integer m consisting of k identical digits and such that m is a multiple of k.

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Mathematica

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A333312 Positive integers m >= 2 such that there is a value of v such that the sequence {v, ..., v + m - 1} of m nonnegative integers can be partitioned into two subsets of consecutive integers with the same sum.

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Python

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A328343 Numbers k such that it is possible to find k consecutive squares whose sum is equal to the sum of two consecutive squares.

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Comments

Examples

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Programs

Mathematica

Python

Extensions

A309417 Number of steps needed to reduce 10^n to zero by subtracting its digital sum.

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