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A137362
Positions at which the truncated square root of triangular numbers is unique.
1
4, 7, 8, 11, 14, 17, 18, 21, 24, 25, 28, 31, 34, 35, 38, 41, 42, 45, 48, 49, 52, 55, 58, 59, 62, 65, 66, 69, 72, 75, 76, 79, 82, 83, 86, 89, 92, 93, 96, 99, 100, 103, 106, 107, 110, 113, 116, 117, 120, 123, 124, 127, 130, 133, 134, 137, 140, 141, 144, 147, 148, 151, 154, 157, 158, 161, 164, 165, 168
OFFSET
1,1
COMMENTS
For any term p of the sequence B(p)=1+B(p-1)=-1+B(p+1), where B(p) = floor(p*(p-1)/2).
For any of others p, one of these equalities is wrong.
The difference between two successive isolated terms of the sequence is always 3 or 7 (a(4)-a(1)=11-4=7, a(5)-a(4)=14-11=3)
The difference between the first or second terms of two successive pairs of the sequence is always 7 or 10 (a(6)-a(3)=17-7=7=a(7)-a(3)=18-8=10, a(9-a(6)=24-17=a(10)-a(7)=25-18=7)
For any n, a(n+13)-a(n) is always equal to 31 or 33. a(14)-a(1)=35-4=31, a(16)-a(3)=41-8=33.
Consider the slowly rising step function A061288 of truncated square roots. It attains unique (non-repeated) values A061288(j)=2,4,5,7,9,11,12,... once, whereas all others (1,3,6,8,10,..) occur at least twice. The values j+1 of the associated indices j=3,6,7,10,13,16 are listed here. - R. J. Mathar, May 05 2008
n such that A161680(n-1) < (A061288(n)-1)^2. - Robert Israel, Jun 05 2017
LINKS
MAPLE
R:= map(n -> floor(sqrt(n*(n-1)/2)), [$1..100]):
select((t -> R[t] > R[t-1] and R[t] < R[t+1], [$2..99]); # Robert Israel, Jun 05 2017
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Philippe Lallouet (philip.lallouet(AT)orange.fr), Apr 26 2008, Jun 06 2008
EXTENSIONS
Edited by R. J. Mathar, May 05 2008
Edited by, and more terms from, Robert Israel, Jun 05 2017
STATUS
approved