A370362
Numbers k such that any two consecutive decimal digits of k^2 differ by 1 after arranging the digits in decreasing order.
Original entry on oeis.org
0, 1, 2, 3, 18, 24, 66, 74, 152, 179, 3678, 3698, 4175, 4616, 5904, 5968, 6596, 7532, 8082, 8559, 9024, 10128, 10278, 11826, 12363, 12543, 12582, 13278, 13434, 13545, 13698, 14442, 14676, 14766, 15681, 15963, 16854, 17529, 17778, 18072, 19023, 19377, 19569, 19629
Offset: 1
18^2 = 324 consists of the consecutive digits 2, 3 and 4;
24^2 = 576 consists of the consecutive digits 5, 6 and 7;
66^2 = 4356 consists of the consecutive digits 3, 4, 5 and 6;
74^2 = 5476 consists of the consecutive digits 4, 5, 6 and 7.
The actual squares are given by
A370610.
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isconsecutive(m, {b=10})=my(v=vecsort(digits(m, b))); for(i=2, #v, if(v[i]!=1+v[i-1], return(0))); 1 \\ isconsecutive(k, b) == 1 if and only if any two consecutive digits of the base-n expansion of m differ by 1 after arranging the digits in decreasing order
a(n) = isconsecutive(n^2)
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from math import isqrt
from sympy.ntheory import digits
def afull(): return([i for i in range(isqrt(10**10)+1) if len(d:=sorted(str(i*i))) == ord(d[-1])-ord(d[0])+1 == len(set(d))])
print(afull()) # Michael S. Branicky, Feb 23 2024
A370370
Number of squares such that any two consecutive digits of their base-n expansions differ by 1 after arranging the digits in decreasing order.
Original entry on oeis.org
2, 2, 6, 3, 10, 12, 14, 48, 160, 148, 226, 54, 1277, 2675, 6812, 2525
Offset: 2
a(4) = 6 because there are 6 such squares in base 4: 0^2 = 0 = 0_4, 1^2 = 1 = 1_4, 2^2 = 4 = 10_4, 3^2 = 9 = 21_4, 6^2 = 36 = 210_4 and 15^2 = 225 = 3201_4.
a(6) = 10 because there are 10 such squares in base 6: 0^2 = 0 = 0_6, 1^2 = 1 = 1_6, 2^2 = 4 = 2_6, 9^2 = 81 = 213_6, 11^2 = 121 = 321_6, 21^2 = 441 = 2013_6, 50^2 = 2500 = 15324_6, 75^2 = 5625 = 42013_6, 85^2 = 7225 = 53241_6 and 195^2 = 38025 = 452013_6.
a(10) = 160 because there are 160 terms in A370362 (or A370610).
Cf.
A258103 (number of pandigital squares in base n).
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isconsecutive(m,n)=my(v=vecsort(digits(m,n))); for(i=2, #v, if(v[i]!=1+v[i-1], return(0))); 1 \\ isconsecutive(k,n) == 1 if and only if any two consecutive digits of the base-n expansion of m differ by 1 after arranging the digits in decreasing order
a(n) = my(lim=sqrtint(if(n%2==1 && valuation(n-1, 2)%2==0, n^(n-1) - (n^(n-1)-1)/(n-1)^2, n^n - (n^n-n)/(n-1)^2)), count=0); for(m=0, lim, if(isconsecutive(m^2,n), count++)); count \\ See A258103 for the searching limit of m
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# replace n**n with ub in A370371 for faster version
from math import isqrt
from sympy.ntheory import digits
def a(n): return(sum(1 for i in range(isqrt(n**n)+1) if len(d:=sorted(digits(i*i, n)[1:])) == d[-1]-d[0]+1 == len(set(d))))
print([a(n) for n in range(2, 12)]) # Michael S. Branicky, Feb 23 2024
A370611
Largest square such that any two consecutive digits of its base-n expansion differ by 1 after arranging the digits in decreasing order.
Original entry on oeis.org
1, 1, 225, 4, 38025, 751689, 10323369, 355624164, 9814072356, 279740499025, 8706730814089, 86847500601, 11027486960232964, 435408094460869201, 18362780530794065025, 2492638430009890761, 39207739576969100808801, 1972312183619434816475625, 104566626183621314286288961, 13215338757299095309775089
Offset: 2
See the Example section of A370371.
The square roots are given by
A370371.
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isconsecutive(m, n)=my(v=vecsort(digits(m, n))); for(i=2, #v, if(v[i]!=1+v[i-1], return(0))); 1 \\ isconsecutive(k, n) == 1 if and only if any two consecutive digits of the base-n expansion of m differ by 1 after arranging the digits in decreasing order
a(n) = forstep(m=sqrtint(if(n%2==1 && valuation(n-1, 2)%2==0, n^(n-1) - (n^(n-1)-1)/(n-1)^2, n^n - (n^n-n)/(n-1)^2)), 0, -1, if(isconsecutive(m^2, n), return(m^2)))
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